Mathematics

# Evaluate$\displaystyle \int_{0}^{\sin ^{2}t}\sin ^{-1}\sqrt{x}dx+\int_{0}^{\cos ^{2}t}\cos ^{-1}\sqrt{x}dx = k$, then $tan(k) =?$

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##### SOLUTION
Let $\displaystyle I=\int _{ 0 }^{ \sin ^{ 2 } t } \sin ^{ -1 } \sqrt { x } dx+\int _{ 0 }^{ \cos ^{ 2 } t } \cos ^{ -1 } \sqrt { x } dx$
$I=I_{ 1 }+I_{ 2 }$
For $I_{ 1 }$ put $\sqrt { x } =\sin \theta \Rightarrow x=\sin ^{ 2 } \theta \Rightarrow dx=\sin 2\theta d\theta$
$\therefore I_{ 1 }=\int _{ 0 }^{ t } \theta \sin 2\theta d\theta$
And for $I_{ 2 }$ put $\sqrt { x } =\cos \phi$
$\therefore I_{ 2 }=\int _{ \pi /2 }^{ t } -\phi \sin 2\phi d\phi =\int _{ t }^{ \pi /2 } \theta \sin 2\theta d\theta$
Hence
$\displaystyle I=\int _{ 0 }^{ t } \theta \sin 2\theta d\theta +\int _{ t }^{ \pi /2 } \theta \sin 2\theta d\theta =\int _{ 0 }^{ \pi /2 } \theta \sin 2\theta d\theta$
$\displaystyle \\ =\left[ \theta \left( -\dfrac { \cos 2\theta }{ 2 } \right) -\left( 1 \right) \left( -\dfrac { \sin 2\theta }{ 4 } \right) \right] _{ 0 }^{ \pi /2 }=\dfrac { \pi }{ 2 } .\dfrac { 1 }{ 2 } +.0=\dfrac { \pi }{ 4 }$

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One Word Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle \int_0^{\pi} \frac{\phi d \phi}{1 + \sin \phi}$ is equal to
• A. $- \pi$
• B. $\displaystyle \frac{\pi}{2}$
• C. None of these
• D. $\pi$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\underset {n\rightarrow \infty}{\lim}\dfrac {1}{n}\displaystyle \sum_{r = 1}^{r = 2n} \dfrac {r}{\sqrt {n^{2} + r^{2}}}$ equals.
• A. $\sqrt {5} + 1$
• B. $1 = \sqrt {5}$
• C. None of these
• D. $\sqrt {5} - 1$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\int_{-3\pi /2}^{-\pi /2}\left [ \left ( x+\pi \right )^{3}+\cos ^{2}\left ( x+3\pi \right ) \right ]dx$ is equal to
• A. $\displaystyle \frac{\pi ^{4}}{32}$
• B. $\displaystyle \frac{\pi ^{4}}{32}+\frac{\pi }{2}$
• C. $\displaystyle \frac{\pi }{2}-1$
• D. $\displaystyle \frac{\pi }{2}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$\int_{\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{sin x +cos x}{\sqrt{sin 2x}}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Single Correct Medium
Evaluate: $\displaystyle \int_{0}^{1} \cos$ $\left(2 \cot^{-1}\sqrt{\displaystyle \frac{1- {x}}{1+ {x}}}\right)dx$
• A. $\dfrac{1}{2}$
• B. $0$
• C. $1$
• D. $\dfrac{-1}{2}$