MHT CET · Maths · Indefinite Integration
\(\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\)
- A \(\frac{-5}{2}[\tan \sqrt{x}]^5+c\)
- B \([\tan \sqrt{\mathrm{x}}]^5+\mathrm{c}\)
- C \(\frac{2}{5}[\tan \sqrt{x}]^5+c\)
- D \(\frac{5}{2}[\tan \sqrt{x}]^5+c\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{5}[\tan \sqrt{x}]^5+c\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \text { Let } I=\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x \\
& \text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\
& \therefore I=2 \int \tan ^4 \operatorname{tsec}^2 t d t
\end{aligned}
\)
Put \(\tan \mathrm{t}=\mathrm{u} \quad \Rightarrow \sec ^2 \mathrm{tdt}=\mathrm{du}\)
\(
\begin{aligned}
& \therefore \mathrm{I}=2 \int \mathrm{u}^4 \mathrm{du} \\
& =\frac{2 \mathrm{u}^5}{5}=\frac{2(\tan \mathrm{t})^5}{5}=\frac{2}{5} \tan ^5 \sqrt{\mathrm{x}}+\mathrm{c}
\end{aligned}
\)
\begin{aligned}
& \text { Let } I=\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x \\
& \text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}} d x=d t \\
& \therefore I=2 \int \tan ^4 \operatorname{tsec}^2 t d t
\end{aligned}
\)
Put \(\tan \mathrm{t}=\mathrm{u} \quad \Rightarrow \sec ^2 \mathrm{tdt}=\mathrm{du}\)
\(
\begin{aligned}
& \therefore \mathrm{I}=2 \int \mathrm{u}^4 \mathrm{du} \\
& =\frac{2 \mathrm{u}^5}{5}=\frac{2(\tan \mathrm{t})^5}{5}=\frac{2}{5} \tan ^5 \sqrt{\mathrm{x}}+\mathrm{c}
\end{aligned}
\)
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