Mathematics

# $\int\limits_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx}$

##### SOLUTION
$\begin{array} { l l } { \text { Let } x = \tan \theta } & { \Longrightarrow \theta = \tan ^ { - 1 } x } \\ { } & { \Longrightarrow \sin ^ { - 1 } \frac { 2 x } { 1 + x ^ { 2 } } = \sin ^ { - 1 } \frac { 2 \tan \theta } { 1 + \tan ^ { 2 } \theta } } \\ { } & { = \sin ^ { - 1 } ( \sin 2 \theta ) \quad \text { using } ( 1 ) } \\ { } & { = 2 \theta = 2 \tan ^ { - 1 } x } \end{array}$
$\begin{array} { l } { \text { Now the integral becomes fairly simple } } \\ { I = 2 \int _ { 1 } ^ { 0 } \tan ^ { - 1 } x \cdot 1 d x } \\ { \text { Integrating by parts } } \\ { \int u \cdot v d x = u \int v d x - \int \left( \frac { d u } { d x } \cdot \int v d x \right) d x } \end{array}$
$\begin{array} { l } { \text { Taking } v = 1 \quad \& \quad u = \tan ^ { - 1 } x } \\ { \Longrightarrow I = x \cdot \tan ^ { - 1 } \left. x \right| _ { 0 } ^ { 1 } - \left. \frac { \ln \left| 1 + x ^ { 2 } \right| } { 2 } \right| _ { 0 } ^ { 1 } } \end{array}$
$\Longrightarrow I = \frac { \pi } { 4 } - \frac { \ln 2 } { 2 }$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
Evaluate $\displaystyle \int_{-1}^{2} |x^{3} - x| dx$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
$\displaystyle \int \sqrt{\left ( \frac{x}{1+x^{3}} \right )}$dx is equal to $\frac{n}{m}\log \left \{ t+\sqrt{p+t^{k}} \right \}.$FInd $k+m+p+n$ ? where $\displaystyle x^{3/2}=t$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
Evaluate $\displaystyle \int (\sqrt {\tan x} + \sqrt {\cot x})dx$.
• A. $\sqrt {2}\tan^{-1} (\sqrt {\tan x} - \sqrt {\cot x}) + C$
• B. $\dfrac {2}{\sqrt {2}}\tan^{-1} x + C$
• C. None of the above
• D. $\sqrt {2}\tan^{-1} \left (\dfrac {\sqrt {\tan x} - \sqrt {\cot x}}{\sqrt {2}} \right ) + C$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate $\displaystyle\int\dfrac{e^{-x}}{1+e^x}dx$

1 Verified Answer | Published on 17th 09, 2020

Q5 Passage Hard
If $n\rightarrow \infty$ then the limit of series in $n$ can be evaluated by following the rule : $\displaystyle \lim_{n\rightarrow \infty}\sum_{r=an+b}^{cn+d}\frac{1}{n}f\left ( \frac{r}{n} \right )=\int_{a}^{c}f(x)dx,$
where in $LHS$, $\dfrac{r}{n}$ is replaced by $x$,
$\dfrac{1}{n}$ by $dx$
and the lower and upper limits are $\lim_{n\rightarrow \infty }\dfrac{an+b}{n}\, and \, \lim_{n\rightarrow \infty }\dfrac{cn+d}{n}$ respectively.