Clase 12 Soluciones RD Sharma – Capítulo 20 Integrales definidas – Ejercicio 20.1

Evalúe las siguientes integrales definidas:

Pregunta 23. $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$

yo = $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2(1-cos^2x))dx$

yo = $\int_{0}^{\frac{\pi}{2}} [(a^2-b^2)cos^2x+b^2]dx$

yo = $\int_{0}^{\frac{\pi}{2}} [(a^2-b^2)(\frac{1+cos2x}{2})]dx+\int_{0}^{\frac{\pi}{2}}b^2dx$

yo = $\frac{a^2-b^2}{2}\int_{0}^{\frac{\pi}{2}} [(1+cos2x)]dx+\int_{0}^{\frac{\pi}{2}}b^2dx$

yo = $\frac{a^2-b^2}{2}\left[x+\frac{sin2x}{2}\right]_{0}^{\frac{\pi}{2}}+b^2\left[x\right]_{0}^{\frac{\pi}{2}}$

yo = $\frac{a^2-b^2}{2}\left[\frac{\pi}{2}+sin\pi-0-0\right]+b^2\left[\frac{\pi}{2}-0\right]$

yo = $\frac{a^2-b^2}{2}\left[\frac{\pi}{2}\right]+b^2\left[\frac{\pi}{2}\right]$

yo = $(\frac{a^2-b^2}{2}+b^2)\left[\frac{\pi}{2}\right]$

yo = $(\frac{a^2-b^2+2b^2}{2})\left[\frac{\pi}{2}\right]$

yo = [(a ² + b ² )/2][π/2]

yo = π(a ² + b ² )/4

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$ es π(a ² + b ² )/4.

Pregunta 24. $\int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{1+\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{\frac{1+tan^2\frac{x}{2}+2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{1+tan^2\frac{x}{2}}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(1+tan\frac{x}{2})^2}{sec^2\frac{x}{2}}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \frac{1+tan\frac{x}{2}}{sec\frac{x}{2}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} (cos\frac{x}{2}+sin\frac{x}{2})dx$

yo = $\left[2sin\frac{x}{2}-2cos\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}$

yo = 2[senπ/4 – cosπ/4 – 0 + 1]

Yo = 2[1/√2 – 1/√2 – 0 + 1]

yo = 2 (1)

yo = 2

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx$ es 2.

Pregunta 25. $\int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{2cos^2\frac{x}{2}}dx$

yo = $\int_{0}^{\frac{\pi}{2}} \sqrt{2}cos\frac{x}{2}dx$

yo = $\sqrt{2}\int_{0}^{\frac{\pi}{2}}cos\frac{x}{2}dx$

yo = $\sqrt{2}\left[2sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}$

yo = $2\sqrt{2}\left[sin\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}$

yo = 2√2[senπ/4 – sen0]

yo = 2√2[1/√2- sen0]

yo = 2√2[1/√2]

yo = 2

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx$ es 2.

Pregunta 26. $\int_{0}^{\frac{\pi}{2}} xsinxdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} xsinxdx$

Al usar la integración por partes, obtenemos,

yo = x ∫senxdx – ∫(∫sen x (1)dx)dx

yo = -xcosx – ∫(∫sen xdx)dx

yo = -xcosx + ∫cosxdx

I = -xcosx + senx

Entonces obtenemos,

yo = $\left[-xcosx+sinx\right]^{\frac{\pi}{2}}_0$

yo = [-π/2cosπ/2 + senπ/2 + 0 – 0]

yo = 0 + 1 + 0 – 0

yo = 1

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} xsinxdx$ es 1.

Pregunta 27. $\int_{0}^{\frac{\pi}{2}} xcosxdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} xcosxdx$

Al usar la integración por partes, obtenemos,

yo = x∫cosxdx – ∫(∫cos x (1)dx)dx

yo = xsenx – ∫(∫cosxdx)dx

I = xsenx – ∫senxdx

I = x sen x + cos x

Entonces obtenemos,

yo = $\left[xsinx+cosx\right]^{\frac{\pi}{2}}_0$

yo = [π/2senπ/2 + cosπ/2 – 0 – cos0]

yo = π/2 + 0 – 0 – 1

yo = π/2 – 1

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} xcosxdx$ es π/2 – 1.

Pregunta 28. $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$

Al usar la integración por partes, obtenemos,

yo = x ² senx – ∫(2x∫(cosx)dx)dx

yo = x ² senx – ∫(2xsenx)dx

yo = x ² senx – 2[-xcosx – ∫(1∫sinxdx)dx]

I = x ² senx – 2[-xcosx + ∫sinxdx]

I = x ² senx – 2[-xcosx + senx]

yo = x ² senx + 2xcosx – 2senx

Entonces obtenemos,

yo = $\left[x^2sinx+2xcosx-2sinx\right]^{\frac{\pi}{2}}_0$

yo = [(π/2) ² sinπ/2 + 2(π/2)cosπ/2 – 2sinπ/2 – 0 – 0 + sin0]

yo = [π ² /4 + 0 – 2 – 0 – 0 + 0]

yo = π ² /4 – 2

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$ es π ² /4 – 2.

Pregunta 29. $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$

Al usar la integración por partes, obtenemos,

yo = -x ² cosx – ∫(2x∫senxdx)dx

yo = -x ² cosx + ∫(2xcosx)dx

yo = -x ² cosx + 2[xsenx – ∫(∫cosxdx)dx]

I = -x ² cosx + 2[xsenx – ∫senxdx]

I = -x ² cosx + 2[xsenx + cosx]

yo = -x ² cosx + 2xsenx + 2cosx

Entonces obtenemos,

yo = $\left[-x^2cosx+2xsinx+2cosx\right]^{\frac{\pi}{4}}_0$

yo = -(4) ² cos4 + 2π/4senπ/4 + 2cosπ/4 + 0 – 0 – 2

yo = – ² 16(1/√2) + π/2(1/√2) + 2(1/√2) + 0 – 0 – 2

yo = – ² 16√2 + π/2√2 + √2 – 2

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$ es -π ² /16√2 + π/2√2 + √2 – 2.

Pregunta 30. $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$

Al usar la integración por partes, obtenemos,

yo = 1/2x ² sen2x – ∫(2x∫cos2xdx)dx

yo = 1/2x ² sen2x – ∫(xsen2x)dx

yo = 1/2x ² sen2x – [-1/2xcos2x – ∫(∫sen2xdx)dx]

I = 1/2x ² sen2x – [-1/2xcos2x + ∫1/2 cos2xdx]

I = 1/2x ² sen2x – [-1/2xcos2x + 1/4sen2xdx]

yo = 1/2x ² sen2x + 1/2xcos2x – 1/4sen2xdx

Entonces obtenemos,

yo = $\left[\frac{1}{2}x^2sin2x+\frac{1}{2}xcos2x-\frac{1}{4}sin2xdx\right]^{\frac{\pi}{2}}_0$

yo = [1/2(π ² /4)senπ + 1/2(π/2)cosπ – 0 – 0 – 0 + 0]

yo = -π/4

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$ es -π/4.

Pregunta 31. $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$

Solución:

Tenemos,

yo = $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$

yo = $\int_{0}^{\frac{\pi}{2}} x^2(\frac{1+cos2x}{2})dx$

yo = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2+x^2cos2x)dx$

yo = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}} x^2dx+\frac{1}{2}\int_{0}^{\frac{\pi}{2}} (x^2cos2x)dx$

Al usar la integración por partes, obtenemos,

I = 1/2[x ³ /3] + x ² sin2x/2 – [x ∫sin2x – ∫(∫sin2xdx)dx]

I = 1/2[x ³ /3] + x ² sen2x/2 + xcosx/2 – sen2x/4

Entonces obtenemos,

yo = $\left[\frac{1}{2}[\frac{x^3}{3}]+\frac{x^2sin2x}{2}+\frac{xcosx}{2}-\frac{sin2x}{4}\right]^{\frac{\pi}{2}}_0$

yo = [1/6[π 3/8 ^] + 0 + 0 – π/8]

yo = π ^3/48 – π/8

Por lo tanto, el valor de $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$ es π ^3/48 – π/8.

Pregunta 32. $\int_{1}^{2}logxdx$

Solución:

Tenemos,

yo = $\int_{1}^{2}logxdx$

Al usar la integración por partes, obtenemos,

yo = $xlogx(1)-\int x\frac{1}{x}dx$

yo = xlogx – ∫1dx

yo = xlogx – x

Entonces obtenemos,

yo = $\left[xlogx-x\right]^2_1$

I = 2log2 – 2 – log1 + 1

I = 2 log 2 – 1

Por lo tanto, el valor de $\int_{1}^{2}logxdx$ es 2 log 2 – 1.

Pregunta 33. $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$

Solución:

Tenemos,

yo = $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$

Al usar la integración por partes, obtenemos,

yo = $(logx)\frac{(x+1)^{-2+1}}{-2+1}-\int (\frac{1}{x}\int \frac{1}{(x+1)^2}dx)dx$

yo = $-(x+1)^{-1}logx+\int \frac{1}{x(x+1)}dx$

yo = $-\frac{logx}{x+1}+\int (\frac{1}{x}-\frac{1}{x+1})dx$

yo = $-\frac{logx}{x+1}+logx - log(x+1)$

Entonces obtenemos,

yo = $\left[-\frac{logx}{x+1}+logx - log(x+1)\right]^3_1$

I = -log3/4 + log3 – log4 + log1/2 – log1 + log2

I = log3(1 – 1/4) – 2log2 + 0 – 0 + log2

I = 3/4 log3 – log2

Por lo tanto, el valor de $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$ es 3/4log3 – log2.

Pregunta 34. $\int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx$

Solución:

Tenemos,

yo = $\int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx$

yo = $\int_{1}^{e}(\frac{e^x}{x}+e^xlogx)dx$

yo = $\int_{1}^{e}\frac{e^x}{x}dx+\int_{1}^{e}e^xlogxdx$

Al usar la integración por partes, obtenemos,

yo = $e^xlogx-\int_{1}^{e}e^xlogxdx+\int_{1}^{e}e^xlogxdx$

yo = e ^x logx

Entonces obtenemos,

yo = $\left[e^xlogx\right]^e_1$

I = e ^e loge – e ¹ log1

yo = mi mi ⁽ 1) – 0

yo = ^mi

Por lo tanto, el valor de $\int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx$ es e ^e .

Pregunta 35. $\int_{1}^{e}\frac{logx}{x}dx$

Solución:

Tenemos,

yo = $\int_{1}^{e}\frac{logx}{x}dx$

Sea log x = t, entonces tenemos,

=> (1/x) dx = dt

Ahora, el límite inferior es, x = 1

=> t = log x

=> t = registro 1

=> t = 0

Además, el límite superior es, x = e

=> t = log x

=> t = log e

=> t = 1

Entonces, la ecuación se convierte en,

yo = $\int_{1}^{e}\frac{logx}{x}dx$

yo = $\int_{0}^{1}tdt$

yo = $\left[\frac{t^2}{2}\right]^1_0$

yo = 1/2 – 0/2

yo = 1/2

Por lo tanto, el valor de $\int_{1}^{e}\frac{logx}{x}dx$ es 1/2.

Pregunta 36. $\int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx$

Solución:

Tenemos,

yo = $\int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx$

yo = $\int_{e}^{e^2}\frac{1}{logx}dx-\int_{e}^{e^2}\frac{1}{(logx)^2}dx$

Al usar la integración por partes, obtenemos,

yo = $\frac{x}{logx}-\int [(\frac{-1}{(logx)^2})(\frac{1}{x})\int dx]dx -\int \frac{1}{(logx)^2}dx$

yo = $\frac{x}{logx}-\int \frac{-1}{(logx)^2}dx -\int \frac{1}{(logx)^2}dx$

yo = $\frac{x}{logx}+\int \frac{1}{(logx)^2}dx -\int \frac{1}{(logx)^2}dx$

yo = x/logx

Entonces obtenemos,

yo = $\left[\frac{x}{logx}\right]^{e^2}_e$

yo = $\left[\frac{e^2}{loge^2}-\frac{e}{loge}\right]$

yo = $\left[\frac{e^2}{2loge}-e\right]$

yo = mi ² /2 – mi

Por lo tanto, el valor de $\int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx$ es e ² /2 – e.

Pregunta 37. $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$

Solución:

Tenemos,

yo = $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$

yo = $\int_{1}^{2}\frac{x}{x(x+2)}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx$

yo = $\int_{1}^{2}\frac{1}{x+2}dx+\int_{1}^{2}\frac{3}{x(x+2)}dx$

yo = $\int_{1}^{2}\frac{1}{x+2}dx+\frac{3}{2}\int_{1}^{2}(\frac{1}{x}-\frac{1}{x+2})dx$

yo = $\left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1$

yo = $\left[log(x+2)\right]^2_1+\left[\frac{3}{2}logx-\frac{3}{2}log(x+2)\right]^2_1$

yo = $\left[\frac{3}{2}logx-\frac{1}{2}log(x+2)\right]^2_1$

I = 1/2[3log2 – log4 + log3]

I = 1/2[3log2 – 2log2 + log3]

I = 1/2[registro 2 – registro 3]

yo = 1/2[log6]

yo = log6/2

Por lo tanto, el valor de $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$ es log6/2.

Pregunta 38. $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$

Solución:

Tenemos,

yo = $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{5(2x+3)}{5x^2+1}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{10x+15}{5x^2+1}dx$

yo = $\frac{1}{5}\int_{0}^{1}(\frac{10x}{5x^2+1}+\frac{15}{5x^2+1}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{1}{5}\int_{0}^{1}\frac{15}{5x^2+1}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+3\int_{0}^{1}\frac{1}{5(x^2+\frac{1}{5})}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx$

yo = $\frac{1}{5}\int_{0}^{1}\frac{10x}{5x^2+1}+\frac{3}{5}\int_{0}^{1}\frac{1}{x^2+\frac{1}{5}}dx$

yo = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{5}(\frac{1}{\frac{1}{\sqrt{5}}})tan^{-1}\frac{x}{\frac{1}{\sqrt{5}}}\right]^1_0$

yo = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0$

yo = $\frac{1}{5}\left[log(5x^2+1)\right]^1_0+\left[\frac{3}{\sqrt{5}}tan^{-1}\sqrt{5}x\right]^1_0$

I = [1/5log6 + 3/√5tan ^-1 (√5) – 1/5log1 – 3/√5tan ^-1 (0)]

Yo = [1/5 log6 + 3√5 tan ^-1 (√5) – 0 – 0]

yo = 1/5 log6 + 3√5 tan ^-1 (√5)

Por lo tanto, el valor de $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$ es 1/5 log6 + 3√5 tan ^-1 (√5).

Pregunta 39. $\int_{0}^{2}\frac{1}{x+4-x^2}dx$

Solución:

Tenemos,

yo = $\int_{0}^{2}\frac{1}{x+4-x^2}dx$

yo = $\int_{0}^{2}\frac{1}{-(x^2+x-4)}dx$

yo = $\int_{0}^{2}\frac{1}{-(x^2+x+\frac{1}{4}-4-\frac{1}{4})}dx$

yo = $\int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-\frac{17}{4}}dx$

yo = $\int_{0}^{2}\frac{-1}{(x-\frac{1}{2})^2-(\frac{\sqrt{17}}{2})^2}dx$

yo = $\int_{0}^{2}\frac{1}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}dx$

Sea x – 1/2 = t, entonces tenemos,

=> dx = dt

Ahora, el límite inferior es, x = 0

=> t = x – 1/2

=> t = 0 – 1/2

=> t = 1/2

Además, el límite superior es, x = 2

=> t = x – 1/2

=> t = 2 – 1/2

=> t = 3/2

Entonces, la ecuación se convierte en,

yo = $\int_{\frac{-1}{2}}^{\frac{3}{2}}\frac{1}{(\frac{\sqrt{17}}{2})^2-t^2}dt$

yo = $\left[\frac{1}{2(\frac{\sqrt{17}}{2})}log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\right]^{\frac{3}{2}}_{\frac{-1}{2}}$

yo = $\frac{1}{\sqrt{17}}\left[log\frac{\frac{\sqrt{17}}{2}+\frac{3}{2}}{\frac{\sqrt{17}}{2}-\frac{3}{2}}-log\frac{\frac{\sqrt{17}}{2}-\frac{1}{2}}{\frac{\sqrt{17}}{2}+\frac{1}{2}}\right]$

yo = $\frac{1}{\sqrt{17}}\left[log\frac{\sqrt{17}+3}{\sqrt{17}-3}-log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]$

yo = $\frac{1}{\sqrt{17}}\left[log(\frac{\sqrt{17}+3}{\sqrt{17}-3}×\frac{\sqrt{17}+1}{\sqrt{17}-1})\right]$

yo = $\frac{1}{\sqrt{17}}\left[\log\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}}\right]$

yo = $\frac{1}{\sqrt{17}}\left[\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right]$

yo = $\frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}$

yo = $\frac{1}{\sqrt{17}}\log\frac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}$

yo = $\frac{1}{\sqrt{17}}\log\frac{25+17+10\sqrt{17}}{8}$

yo = $\frac{1}{\sqrt{17}}\log\frac{42+10\sqrt{17}}{8}$

yo = $\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$

Por lo tanto, el valor de $\int_{0}^{2}\frac{1}{x+4-x^2}dx$ es $\frac{1}{\sqrt{17}}\log\frac{21+5\sqrt{17}}{4}$ .

Pregunta 40. $\int_{0}^{1}\frac{1}{2x^2+x+1}dx$

Solución:

Tenemos,

yo = $\int_{0}^{1}\frac{1}{2x^2+x+1}dx$

yo = $\frac{1}{2}\int_{0}^{1}\frac{1}{x^2+\frac{x}{2}+\frac{1}{2}}dx$

yo = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{1}{2}-\frac{1}{16}}dx$

yo = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+\frac{7}{16}}dx$

yo = $\frac{1}{2}\int_{0}^{1}\frac{1}{(x+\frac{1}{4})^2+(\frac{\sqrt{7}}{4})^2}dx$

yo = $\left[\frac{1}{2}\frac{4}{\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0$

yo = $\left[\frac{4}{2\sqrt{7}}\tan^{-1}(\frac{x+\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]^1_0$

yo = $\frac{4}{2\sqrt{7}}\left[\tan^{-1}(\frac{\frac{5}{4}}{\frac{\sqrt{7}}{4}})-\tan^{-1}(\frac{\frac{1}{4}}{\frac{\sqrt{7}}{4}})\right]$

Yo = 4/2√7[tan ^-1 (5/√7) – tan ^-1 (1/√7)]

Yo = 2/√7[tan ^-1 (5/√7) – tan ^-1 (1/√7)]

Por lo tanto, el valor de $\int_{0}^{1}\frac{1}{2x^2+x+1}dx$ es 2/√7[tan ^-1 (5/√7) – tan ^-1 (1/√7)].

Pregunta 41. $\int_{0}^{1}\sqrt{x(1-x)}dx$

Solución:

Tenemos,

yo = $\int_{0}^{1}\sqrt{x(1-x)}dx$

Sea x = sen ² t, entonces tenemos,

=> dx = 2 sen t cos t dt

Ahora, el límite inferior es, x = 0

=> sen ² t = 0

=> sen t = 0

=> t = 0

Además, el límite superior es, x = 1

=> sen ² t = 1

=> sen t = 1

=> t = π/2

Entonces, la ecuación se convierte en,

yo = $\int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(1-sin^2t)}(2sintcost)dt$

yo = $\int_{0}^{\frac{\pi}{2}}\sqrt{sin^2t(cos^2t)}(2sintcost)dt$

yo = $\int_{0}^{\frac{\pi}{2}}(sintcost)(2sintcost)dt$

yo = $\int_{0}^{\frac{\pi}{2}}(2sin^2tcos^2t)dt$

yo = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(4sin^2tcos^2t)dt$

yo = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(sin^22t)dt$

yo = $\frac{1}{2}\int_{0}^{\frac{\pi}{2}}(\frac{1-cos4t}{2})dt$

yo = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos4t)dt$

yo = $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}dt-\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(cos4t)dt$

yo = $\frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{4}\left[\frac{sin4t}{4}\right]^{\frac{\pi}{2}}_0$

yo = $\frac{1}{4}\left[t\right]^{\frac{\pi}{2}}_0-\frac{1}{16}\left[sin4t\right]^{\frac{\pi}{2}}_0$

Yo = 1/4[π/2 – 0] – 1/16[sen2π – 0]

Yo = 1/4[π/2] – 1/16[0 – 0 ]

yo = π/8

Por lo tanto, el valor de $\int_{0}^{1}\sqrt{x(1-x)}dx$ es π/8.

Pregunta 42. $\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx$

Solución:

Tenemos,

yo = $\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx$

yo = $\int_{0}^{2}\frac{1}{\sqrt{3+1-(x^2-2x+1)}}dx$

yo = $\int_{0}^{2}\frac{1}{\sqrt{4-(x^2-2x+1)}}dx$

yo = $\int_{0}^{2}\frac{1}{\sqrt{(2)^2-(x-1)^2}}dx$

yo = $\left[sin^{-1}(\frac{x-1}{2})\right]_{0}^{2}$

I = [sin ^-1 (1/2) – sin ^-1 (-1/2)]

yo = π/6 -(-π/6)

yo = π/6 + π/6

yo = π/3

Por lo tanto, el valor de $\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx$ es π/3 .

Pregunta 43. $\int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx$

Solución:

Tenemos,

yo = $\int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx$

yo = $\int_{0}^{4}\frac{1}{\sqrt{4-4+4x-x^2}}dx$

yo = $\int_{0}^{4}\frac{1}{\sqrt{4-(x^2-4x+4)}}dx$

yo = $\int_{0}^{4}\frac{1}{\sqrt{2^2-(x-2)^2}}dx$

yo = $\left[sin^{-1}(\frac{x-2}{2})\right]^4_0$

yo = $\left[sin^{-1}(\frac{4-2}{2})-sin^{-1}(\frac{0-2}{2})\right]$

I = [sin ^-1 (2/2) – sin ^-1 (-2/2)]

I = sen ^-1 1 – sen ^-1 (-1)

yo = π/2 – (-π/2)

yo = π/2 + π/2

yo = π

Por lo tanto, el valor de $\int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}dx$ es π.

Pregunta 44. $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$

Solución:

Tenemos,

yo = $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$

yo = $\int_{-1}^{1}\frac{1}{x^2+2x+1+4}dx$

yo = $\int_{-1}^{1}\frac{1}{(x+1)^2+2^2}dx$

Sea x + 1 = t, entonces tenemos,

=> dx = dt

Ahora, el límite inferior es, x = –1

=> t = x + 1

=> t = – 1 + 1

=> t = 0

Además, el límite superior es, x = 1

=> t = x + 1

=> t = 1 + 1

=> t = 2

Entonces, la ecuación se convierte en,

yo = $\int_{0}^{2}\frac{1}{t^2+2^2}dt$

yo = $\left[\frac{1}{2}tan^{-1}\frac{t}{2}\right]^2_0$

I = 1/2 bronceado ^-1 2/2 – 1/2 bronceado ^-1 0/2

I = 1/2 bronceado ^-1 1 – 1/2 bronceado ^-1 0

yo = 1/2(π/4) – 0

yo = π/8

Por lo tanto, el valor de $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$ es π/8.

Publicación traducida automáticamente

Artículo escrito por gurjotloveparmar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Clase 12 Soluciones RD Sharma – Capítulo 20 Integrales definidas – Ejercicio 20.1 | conjunto 2

Evalúe las siguientes integrales definidas:

Pregunta 23. $\int_{0}^{\frac{\pi}{2}} (a^2cos^2x+b^2sin^2x)dx$

Pregunta 24. $\int_{0}^{\frac{\pi}{2}} \sqrt{1+sinx}dx$

Pregunta 25. $\int_{0}^{\frac{\pi}{2}} \sqrt{1+cosx}dx$

Pregunta 26. $\int_{0}^{\frac{\pi}{2}} xsinxdx$

Pregunta 27. $\int_{0}^{\frac{\pi}{2}} xcosxdx$

Pregunta 28. $\int_{0}^{\frac{\pi}{2}} x^2cosxdx$

Pregunta 29. $\int_{0}^{\frac{\pi}{4}} x^2sinxdx$

Pregunta 30. $\int_{0}^{\frac{\pi}{2}} x^2cos2xdx$

Pregunta 31. $\int_{0}^{\frac{\pi}{2}} x^2cos^2xdx$

Pregunta 32. $\int_{1}^{2}logxdx$

Pregunta 33. $\int_{1}^{3}\frac{logx}{(x+1)^2}dx$

Pregunta 34. $\int_{1}^{e}\frac{e^x}{x}(1+xlogx)dx$

Pregunta 35. $\int_{1}^{e}\frac{logx}{x}dx$

Pregunta 36. $\int_{e}^{e^2}(\frac{1}{logx}-\frac{1}{(logx)^2})dx$

Pregunta 37. $\int_{1}^{2}\frac{x+3}{x(x+2)}dx$

Pregunta 38. $\int_{0}^{1}\frac{2x+3}{5x^2+1}dx$

Pregunta 39. $\int_{0}^{2}\frac{1}{x+4-x^2}dx$

Pregunta 40. $\int_{0}^{1}\frac{1}{2x^2+x+1}dx$

Pregunta 41. $\int_{0}^{1}\sqrt{x(1-x)}dx$

Pregunta 42. $\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}dx$

Pregunta 44. $\int_{-1}^{1}\frac{1}{x^2+2x+5}dx$

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