Mathematics

# Integrate:$\displaystyle \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

##### SOLUTION

Consider the given integral.

$I=\int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

Let $t={{\sin }^{-1}}x$

$dt=\dfrac{dx}{\sqrt{1-{{x}^{2}}}}$

Therefore,

$I=\int{tdt}$

$I=\dfrac{{{t}^{2}}}{2}+C$

On putting the value of $t$, we get

$I=\dfrac{{{\left( {{\sin }^{-1}}x \right)}^{2}}}{2}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle A: \int \tan^{4}x \sec^{2}xdx=\frac{1}{5}\tan^{5}x+c$
$\displaystyle R: \int [f(x)]^{n}f'(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c$
• A. Both A and R are true but R is not correct explanation of A
• B. A is true R is false
• C. A is false but R is true.
• D. Both A and R are true and R is the correct explanation of A

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate:
$\int\limits_{\pi /4}^{\pi /2} {\sqrt {1 - \sin 2x} } dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\displaystyle \int \frac{\sin \sqrt{x}}{\sqrt{\left ( x \right )}}dx$=
• A. $2\cos \sqrt{x}+c$
• B. $-2\sec \sqrt{x}+c$
• C. $2\sec \sqrt{x}+c$
• D. $-2\cos \sqrt{x}+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Find $\int\limits_0^{\sqrt 2 } {\sqrt {2 - {x^2}} dx}$
• A. 8
• B.
• C. 2
• D. $\frac{\pi }{2}$

Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$