Integrals Test - 48

$$\displaystyle I=\int_{\frac{5}{2}}^{5}\frac{\sqrt{(25-x^2)^3}}{x^4}dx$$
Let $$x= 5\sin\theta$$
$$\displaystyle \therefore dx= 5 \cos\theta d\theta$$
$$\displaystyle \therefore I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{5^3\cos ^3\theta.5\cos\theta }{5^4 \sin^4\theta}d\theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta(\ cosec^2\theta -1)d\theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2

\theta\ cosec^2\theta

d\theta-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta d \theta$$
$$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2

\theta\ cosec^2\theta d\theta- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\

cosec^2 \theta-1)$$
$$\displaystyle =\left [ -\frac{\cot^3\theta}{3} +\cot\theta+\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$$
$$\displaystyle =-0+0+\frac{\pi}{2}-\left ( -\frac{3\sqrt{3}}{3}+\sqrt{3}+\frac{\pi}{6} \right )$$ $$\displaystyle =\frac{\pi}{3}$$

$$\int_{0}^{1}\dfrac{3^{x+1}-4^{x-1}}{12^{x}} dx$$ $$=\displaystyle \int_{0}^{1}(\dfrac{3}{4^{\mathrm{x}}}-\dfrac{1}{4}. \displaystyle \dfrac{1}{3^{\mathrm{x}}}) dx$$

$$=\left \{ 3\dfrac{\dfrac{1}{4^{x}}}{(log\dfrac{1}{4})}-\dfrac{1}{4} \dfrac{\dfrac{1}{3^{x}}}{log\left ( \dfrac{1}{3} \right )}\right \}_{0}^{1}$$

$$=\displaystyle \dfrac{3(\dfrac{1}{4}-1)}{-\log_{\mathrm{e}}4}-\dfrac{1}{4}\dfrac{(\dfrac{1}{3}-1)}{-\log_{\mathrm{e}}3}$$

$$=\displaystyle \dfrac{9}{4}\log_{4}\mathrm{e}-\dfrac{1}{6}\log_{3}\mathrm{e}$$

Ans: B

$$ \displaystyle I= \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$$
Putting $$x=\tan \theta$$ or $$dx= \sec^2 \theta d\theta$$, we get
$$I\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{\theta}{\tan \theta}\sec^2 \theta d \theta$$
$$\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{2\theta}{\sin 2\theta}d\theta$$
Putting  $$2\theta=t$$,i.e.,$$2d\theta=dt$$, we get
$$I\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{t}{\sin t}dt$$
$$\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$$

Putting $$e^x-1=t^2$$ in the given integral, we have
$$ \displaystyle \int_{0

}^{\log

5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=2\int_{0}^{2}\frac{t^2}{t^2+4}dt=2\left

( \int_{0}^{2} 1 dt -4\int_{0}^{2}\frac{dt}{t^2+4}\right )$$
$$ \displaystyle =2\left [ \left ( t-2\tan^{-1}\left ( \frac{t}{2} \right ) \right )_0^2 \right ]$$
$$\displaystyle =2[(2-2\times \frac{\pi}{4})]=4-\pi$$

Orl putting $$x= \sin \theta$$ ,we get $$dx =\cos \theta d \theta$$
Integral (without limits)$$\displaystyle =\int \dfrac{\cos\theta d\theta}{(1+ \sin^2\theta )(\cos \theta)}$$ 

$$\displaystyle =\int \dfrac{d\theta}{1+ \sin^2\theta}=\int\dfrac{\ cosec{^2}\theta d \theta  }{2+\cot^2}$$

$$\displaystyle=\int \dfrac{-dt}{2+t^2}$$ where $$t=\cot \theta$$

$$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}$$

 $$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{\cot\theta}{\sqrt{2}}$$

$$=-\dfrac{1}{\sqrt{2}} \tan^{-1}\dfrac{1}{\sqrt{2}}\left ( \dfrac{\sqrt{1-x^2}}{x} \right )$$

$$\therefore$$ Definite integral $$= -\dfrac{1}{\sqrt{2}}\tan^{-1}1+\dfrac{1}{\sqrt{2}}\tan^{-1}{\infty }$$

$$=-\dfrac{\pi}{4\sqrt{2}}+\dfrac{\pi}{2\sqrt{2}}=\dfrac{\pi}{4\sqrt{2}}$$

In $$\displaystyle{ I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ { x }^{ 3 } }\left( 2-{ x }^{ 3 } \right)  }  } $$
Substitute $$1-{ x }^{ 3 }=t\Rightarrow -3{ x }^{ 2 }dx=dt$$
We get $$\displaystyle{ I }_{ 2 }=-\cfrac { 1 }{ 3 } \int _{ 1 }^{ 0 }{ \cfrac { dt }{ { e }^{ 1-t }\left( 1+t \right)  }  } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t }dt }{ 1+t }  } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x }  } $$
As $$\displaystyle{ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x }  } $$
Therefore $$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =3e$$

$$\displaystyle I=\int _{ -\pi  }^{ \pi  }{ { \left( \cos ax-\sin bx \right)  }^{ 2 }dx } \ =\int _{ -\pi  }^{ \pi  }{ { \sin }^{ 2 }bx\>dx } -2\int _{ -\pi  }^{ \pi  }{ \left( \left( \cos ax \right) \left( \sin bx \right)  \right) dx } +\int _{ -\pi  }^{ \pi  }{ { \cos }^{ 2 }ax\>dx } $$
Let $$\displaystyle I={ I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }$$    ...(1)
Such that, $$\displaystyle{ I }_{ 1 }=\int _{ -\pi  }^{ \pi  }{ { \sin }^{ 2 }bxdx } $$
Substituting $$bx=t\Rightarrow b\>dx=dt$$
We get, $$\displaystyle{ I }_{ 1 }=\cfrac { 1 }{ b } \int _{ -\pi b }^{ \pi b }{ { \sin }^{ 2 }tdt } =\cfrac { 1 }{ b } \int _{ -\pi b }^{ \pi b }{ \left( \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 2 } \cos2t \right) dt } $$
$$\displaystyle=\cfrac { 1 }{ b } { \left[ \cfrac { t }{ 2 } -\cfrac { \sin2t }{ 4 }  \right]  }_{ -\pi b }^{ \pi b }=\pi -\cfrac { \sin\left( 2\pi b \right)  }{ 2b } $$   ...(2)
$$\displaystyle{ I }_{ 2 }=-2\int _{ -\pi  }^{ \pi  }{ \left( \left( \cos ax \right) \left( \sin bx \right)  \right) dx } $$
As $$\left( \cos ax \right) \left( \sin bx \right) $$ is an odd function,
$${ I }_{ 2 }=0$$   ...(3)
$$\displaystyle{ I }_{ 3 }=\int _{ -\pi  }^{ \pi  }{ { \cos }^{ 2 }axdx } $$
Substituting, $$ax=t\Rightarrow adx=dt$$, we get
$$\displaystyle{ I }_{ 3 }=\cfrac { 1 }{ a } \int _{ -\pi a }^{ \pi a }{ { \cos }^{ 2 }tdt } =\cfrac { 1 }{ a } \int _{ -\pi a }^{ \pi a }{ \left( \cfrac { 1 }{ 2 } \cos2t+\cfrac { 1 }{ 2 }  \right) dt } $$
$$\displaystyle=\cfrac { 1 }{ a } { \left[ \cfrac { t }{ 2 } +\cfrac { \sin2t }{ 4 }  \right]  }_{ -\pi a }^{ \pi a }=\cfrac { \sin\left( 2\pi a \right)  }{ 2a } +\pi $$   ...(4)
Substituting (2), (3) and (4) in (1), we get
$$\displaystyle I=\cfrac { \sin\left( 2\pi a \right)  }{ 2a } -\cfrac { \sin\left( 2\pi b \right)  }{ 2b } +2\pi =0+0+2\pi =2\pi $$
Hence, option 'D' is the correct answer.

Let $$I=\int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { x }  } dx } $$

Substitute $$\sqrt { \ln { x }  } =t\Rightarrow \cfrac { dx }{ 2x\sqrt { \ln { x }  }  } =dt\Rightarrow dx=2t{ e }^{ { t }^{ 2 } }dt$$

We get $$I=\int _{ 1 }^{ 2 }{ 2{ t }^{ 2 }{ e }^{ t^{ 2 } }dt } $$
Now integrating by parts, we get

$$I={ \left[ t{ e }^{ t^{ 2 } } \right]  }_{ 1 }^{ 2 }-\int _{ 1 }^{ 2 }{ { e }^{ t^{ 2 } }dt } $$

As $$\int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } }dx } =a$$
Therefore
$$I=2{ e }^{ 4 }-e-a$$
Hence, option 'B' is correct.

Putting $$x =\tan \theta$$,we get
$$\displaystyle \int_{ 0

}^{\infty}\frac{ \:dx}{\left [ x+\sqrt{x^2+1} \right

]^3}=\int_{0}^{\frac{\pi}{2} }\frac{\sec^2\theta d

\theta}{(\tan\theta+\sec\theta)^3}$$
$$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{(1+\sin\theta)^3}d\theta$$
$$\displaystyle =\left [ \frac{1}{2(1+\sin\theta)^2} \right ]_0^{\frac{\pi}{2}}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$$