Mathematics

# $\displaystyle\int x^2\sin^{-1}x dx$.

##### SOLUTION
$\displaystyle \int x^2 \sin^{-1} x dx$
Using byparts,we get
$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{3} \int \cfrac{x^3}{\sqrt{1-x^2}} dx$
Putting $x^2 = m$
$2x dx = dm$
$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x -\cfrac{1}{6} \int \cfrac{m}{\sqrt{1-m }} dm$
$\displaystyle \cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{6} \int \cfrac{1-(1-m)}{\sqrt{1-m}}dm$
$\cfrac{1}{3} x^3 \sin^{-1} x - \cfrac{1}{3} \sqrt{1-x^2} + \cfrac{1}{9} \sqrt{(1-x^2)^3}$

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

#### Realted Questions

Q1 Single Correct Hard
The integral $\displaystyle \int (1+x-\displaystyle \frac{1}{x})e^{x+\frac{1}{x}} dx$ is equal to
• A. $(x-1)e^{x+ \frac{1}{x}} +c$
• B. $(x+1)e^{x+\frac{1}{x}} +c$
• C. $-xe^{x+\frac{1}{x}} +c$
• D. $xe^{x+\frac{1}{x}} +c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
If $f(x)=\begin{vmatrix}\cos{x}&e^{\displaystyle x^2}&\displaystyle 2x\cos^2{\frac{x}{2}}\\x^2&\sec{x}&\sin{x}+x^3\\1&2&x+\tan{x}\end{vmatrix}$, then the value of $\displaystyle\int_{\displaystyle-\frac{\pi}{2}}^{\displaystyle\frac{\pi}{2}}{(x^2+1)(f(x)+f^{\prime\prime}(x))dx}$, is equal to
• A. 1
• B. $-1$
• C. 2
• D. none of these

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Evalue: $\displaystyle \int^{\pi}_{0} \left(sin^2\left(\dfrac{x}{2}\right)-cos^2\left(\dfrac{x}{2}\right)\right)dx$.

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
Evaluate $\displaystyle\int \dfrac{(\sin x)^{2018}}{(\cos x)^{2020}}dx$.
• A. $\dfrac{(\sin x)^{2019}}{2019}+c$
• B. $\dfrac{(\cos x)^{2019}}{2019}+c$
• C. $\dfrac{(\tan x)^{2019} \sec^2 x}{2019}+c$
• D. $\dfrac{(\tan x)^{2019}}{2019}+c$

Let $n \space\epsilon \space N$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1, 2n+2, ...,$ up to $n^{th}$ number are $A_{n}$, $G_{n}$, $H_{n}$ and $R_{n}$ respectively.