MHT CET · Maths · Indefinite Integration
The value of \(\int x \sin x \sec ^{3} x d x\) is
- A \(\frac{1}{2}\left[\sec ^{2} x-\tan x\right]+c\)
- B \(\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c\)
- C \(\frac{1}{2}\left[x \sec ^{2} x+\tan x\right]+c\)
- D \(\frac{1}{2}\left[\sec ^{2} x+\tan x\right]+c\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c\)
Step-by-step Solution
Detailed explanation
\(\int x \sin x \sec ^{3} x d x =\int x \sin x \frac{1}{\cos ^{3} x} d x \)
\( =\int x \tan x \cdot \sec ^{2} x d x \)
\( \text { Put } \tan x=t \Rightarrow \sec ^{2} x d x=d t \)
\( \text { and } \quad x =\tan ^{-1} t\)
Then, it reduces to
\(
\begin{array}{l}
\int \tan ^{-1} t \cdot t d t=\frac{t^{2}}{2} \tan ^{-1} t-\int \frac{t^{2}}{2\left(1+t^{2}\right)} d t \\
=\frac{x \tan ^{2} x}{2}-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t+c \\
=\frac{x\left(\sec ^{2} x-1\right)}{2}-\frac{1}{2} \tan x+\frac{1}{2} x+c \\
=\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c
\end{array}
\)
\( =\int x \tan x \cdot \sec ^{2} x d x \)
\( \text { Put } \tan x=t \Rightarrow \sec ^{2} x d x=d t \)
\( \text { and } \quad x =\tan ^{-1} t\)
Then, it reduces to
\(
\begin{array}{l}
\int \tan ^{-1} t \cdot t d t=\frac{t^{2}}{2} \tan ^{-1} t-\int \frac{t^{2}}{2\left(1+t^{2}\right)} d t \\
=\frac{x \tan ^{2} x}{2}-\frac{1}{2} t+\frac{1}{2} \tan ^{-1} t+c \\
=\frac{x\left(\sec ^{2} x-1\right)}{2}-\frac{1}{2} \tan x+\frac{1}{2} x+c \\
=\frac{1}{2}\left[x \sec ^{2} x-\tan x\right]+c
\end{array}
\)
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