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Match the following items from List I into List II
Select the correct choice
Options:
Select the correct choice
Solution:
2446 Upvotes
Verified Answer
The correct answer is:
$1-\mathrm{C}, 2-\mathrm{D}, 3-\mathrm{B}, 4-\mathrm{A}$
$$
\text { (i) } \int \frac{\sin ^2 x}{\cos ^4 x} d x=\int \tan ^2 x \cdot \sec ^2 x d x
$$
Let $\tan x=t, \sec ^2 x d x=d t$
$$
\int t^2 d t=\frac{t^3}{3}+C=\frac{\tan ^3 x}{3}+C
$$
$$
\begin{aligned}
& \text { (ii) } \int \frac{\sin ^4 x}{\cos ^2 x} d x=\int \frac{\left(1-\cos ^2 x\right)^2}{\cos ^2 x} d x \\
& =\int\left(\sec ^2 x+\cos ^2 x-2\right) d x \\
& =\int \sec ^2 x d x+\int \cos ^2 x d x-\int 2 d x \\
& =\tan x-2 x+\int \frac{1+\cos 2 x}{2} d x \\
& =\tan x-2 x+\frac{1}{2} x+\frac{1}{4} \sin 2 x+C \\
& =\tan x+\frac{\sin 2 x}{4}-\frac{3 x}{2}+C
\end{aligned}
$$
\text { (i) } \int \frac{\sin ^2 x}{\cos ^4 x} d x=\int \tan ^2 x \cdot \sec ^2 x d x
$$
Let $\tan x=t, \sec ^2 x d x=d t$
$$
\int t^2 d t=\frac{t^3}{3}+C=\frac{\tan ^3 x}{3}+C
$$
$$
\begin{aligned}
& \text { (ii) } \int \frac{\sin ^4 x}{\cos ^2 x} d x=\int \frac{\left(1-\cos ^2 x\right)^2}{\cos ^2 x} d x \\
& =\int\left(\sec ^2 x+\cos ^2 x-2\right) d x \\
& =\int \sec ^2 x d x+\int \cos ^2 x d x-\int 2 d x \\
& =\tan x-2 x+\int \frac{1+\cos 2 x}{2} d x \\
& =\tan x-2 x+\frac{1}{2} x+\frac{1}{4} \sin 2 x+C \\
& =\tan x+\frac{\sin 2 x}{4}-\frac{3 x}{2}+C
\end{aligned}
$$
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