AMU MCA Indefinite Integration Previous Year Questions (PYQs)
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Showing 4 questions
$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to
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Solution
$ \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x} $
$ = \sec^2 x - cosec^2 x $
$ \Rightarrow \int (\sec^2 x - cosec^2 x) dx $
$ = \tan x + \cot x + C $
The integral $\int e^x (1+\tan x)\sec x , dx$ equals
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Solution
$ e^x (1+\tan x)\sec x = e^x (\sec x + \sec x \tan x) $
Let $f(x)=\sec x$, then $f'(x)=\sec x \tan x$
$ \Rightarrow \int e^x [f(x)+f'(x)] dx = e^x f(x) + C $
$ = e^x \sec x + C $
$\int \frac{x^3}{1 + x^4} , dx$ equals
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Solution
Let $t = 1 + x^4 \Rightarrow dt = 4x^3 dx$
$\int \frac{x^3}{1 + x^4} dx = \frac{1}{4} \int \frac{dt}{t}$
$= \frac{1}{4}\log(1 + x^4) + C$
$\int \sqrt{16 - 9x^2} , dx$ equals
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Solution
Use standard formula:
$\int \sqrt{a^2 - b^2 x^2} dx = \frac{x}{2}\sqrt{a^2 - b^2 x^2} + \frac{a^2}{2b}\sin^{-1}\left(\frac{bx}{a}\right) + C$
Here $a=4,\ b=3$
$= \frac{x}{2}\sqrt{16 - 9x^2} + \frac{16}{6}\sin^{-1}\left(\frac{3x}{4}\right)$
$= \frac{x}{2}\sqrt{16 - 9x^2} + \frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right) + C$
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