If $\int e^x(1+x) \cdot \sec ^2\left(x e^x\right) d x$ $=f(x)+$ constant, then $f(x)$ is equal to
If $\int e^x(1+x) \cdot \sec ^2\left(x e^x\right) d x$ $=f(x)+$ constant, then $f(x)$ is equal to
- $\cos \left(x e^x\right.$
- $\sin \left(x e^x\right.$
- $2 \tan ^{-1}(x)$
- $\tan \left(x e^x\right)$
Solution
Given that,
$
\int e^x(1+x) \cdot \sec ^2\left(x e^x\right) d x=f(x)+\text { constant }
$
Put
$
x e^x=t \text { in LHS }
$
$
\begin{aligned}
& \Rightarrow & \left(e^x+x e^x\right) d x & =d t \\
& \therefore & \text { LHS } & =\int \sec ^2 t d t \\
& & & =\tan t+\text { constant } \\
& \Rightarrow & \tan \left(x e^x\right)+\text { constant } & =f(x)+\text { constant } \\
& \Rightarrow & f(x) & =\tan \left(x e^x\right)
\end{aligned}
$
Asked in: AP EAMCET 2008
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