Mathematics

# $\int \dfrac {\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}dx=$

$\dfrac {2}{\pi}[x \sin^{-1}x-x \cos^{-1}x+2\sqrt {1-x^{2}}]+c$

##### SOLUTION
$\displaystyle \int{\dfrac{\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}}dx$
$\sin^{-1}x+\cos^{-1}x=\pi /2$
$\dfrac{2}{\pi}\displaystyle \int{\sin^{-1}x}\displaystyle \int{\cos^{-1}x}dx$
$\Rightarrow \dfrac{2}{\pi}\left[x\sin^{-1}x+\sqrt{1-x^2}-x\cos^{-1}x+\sqrt{1-x^2}\right]+c$
$=\dfrac{2}{\pi}[x\sin^{-1}x-x\cos^{-1}x+2\sqrt{1-x^2}]+c$

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

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle \frac{\pi }{4}< \alpha < \displaystyle \frac{\pi }{2}$, value of $\displaystyle \int_{-\pi /2}^{\pi /2}\displaystyle \frac{\sin 2x}{\sqrt{1+\sin 2\alpha \sin x}}$ is
• A. $-\displaystyle \frac{4}{3}\tan \alpha \sec \alpha$
• B. $-\displaystyle \frac{4}{3}\tan \alpha cosec\alpha$
• C. $-\displaystyle \frac{4}{3}\cot \alpha \sec \alpha$
• D. $-\displaystyle \frac{4}{3}\cot \alpha cosec\alpha$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Hard
Evaluate $\displaystyle{\int}^{\pi}_0 \dfrac{x \sin x}{1+\cos^2 x}dx$.

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evaluate $\int _{ 0 }^{ 3 }{ \left( 2{ x }^{ 2 }+3x+5 \right) dx }$ as limit of a sum.

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Hard
Evaluate: $\int_{e}^{e^{2}} \dfrac {dx}{x\log x}.$

Evaluate $\int {\dfrac{{dx}}{{\sqrt {x + 1} \, - \sqrt x }}}$