Mathematics

# Evaluate the definite integral   $\displaystyle \int_0^{\frac {\pi}{2}}\sin 2x \tan^{-1}(\sin x)dx$

##### SOLUTION
Let $I\displaystyle =\int_0^{\frac {\pi}{2}}\sin 2x \tan^{-1}(\sin x)dx=\int_0^{\frac {\pi}{2}}2 \sin x \cos x \tan^{-1}(\sin x)dx$
Also, let $\sin x=t\Rightarrow \cos x dx=dt$
When $x=0, t=0$ and when $x=\dfrac {\pi}{2}, t=1$
$\Rightarrow\displaystyle I=2\int_0^1\tan^{-1}(t)dt$ ............ (1)
Consider $\displaystyle\int t\cdot \tan^{-1}t dt=\tan^{-1}t\cdot \int t dt-\int \left \{\frac {d}{dt}(\tan^{-1}t)\int t dt\right \}dt$
$\displaystyle=\tan^{-1}t\cdot \frac {t^2}{2}-\int \frac {1}{1+t^2}\cdot \frac {t^2}{2}dt$
$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\int \frac {t^2+1-1}{1+t^2}dt$
$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\int 1 dt+\frac {1}{2}\int \frac {1}{1+t^2}dt$
$\displaystyle=\frac {t^2\tan^{-1}t}{2}-\frac {1}{2}\cdot t+\frac {1}{2}\tan^{-1}t$
$\Rightarrow\displaystyle \int_0^1t\cdot \tan^{-1}t dt=\left [\frac {t^2\cdot\tan^{-1}t}{2}-\frac {t}{2}+\frac {1}{2}\tan^{-1}t\right ]_0^1$
$\displaystyle=\frac {1}{2}\left [\frac {\pi}{4}-1+\frac {\pi}{4}\right ]$
$\displaystyle=\frac {1}{2}\left [\frac {\pi}{2}-1\right ]=\frac {\pi}{4}-\frac {1}{2}$
From equation (1), we obtain
$I\displaystyle=2\left [\frac {\pi}{4}-\frac {1}{2}\right ]=\frac {\pi}{2}-1$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\overset{\pi/2}{\underset{0}{\displaystyle\int}}\cos^3xdx =$ ___________.
• A. $\dfrac{8}{3}$
• B. $0$
• C. $1$
• D. $\dfrac{2}{3}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\displaystyle \int_{\pi /4}^{3\pi/4 }\dfrac{dx}{1+\cos x}$ is equal to
• A. $-2$
• B. $4$
• C. $-1$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve: $\int e^{2x}. \sin 3x \,dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
$I = \int {\frac{{\log x}}{{{{\left( {1 + \log x} \right)}^2}}}} dx$

$\int \frac{2x^{2}}{3x^{4}2x} dx$