Integrate the function w.r.t. x: ^ 4 x ^ 2 x x - Vidyadip

Question

Integrate the function w.r.t. x: $\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}$

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Answer

Derivative of $\sqrt{x}$ is $\frac{1}{2} x^{-\frac{1}{2}}=\frac{1}{2 \sqrt{x}}$.
Thus, we use the substitution $\sqrt{x}=t$ so that $\frac{1}{2 \sqrt{x}} d x=d t$ giving $dx = 2t\ dt.$
Thus, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x$ = $\int \frac{2 t \tan ^{4} t \sec ^{2} t d t}{t}$ = $2 \int \tan ^{4} t \sec ^{2} t d t$
Again, we make another substitution $\tan t = u$ so that $\sec^2 t dt = du$
Therefore, $2 \int \tan ^{4} t \sec ^{2} t\ d t=2 \int u^{4} d u=2 \frac{u^{5}}{5}+C$
$= \frac{2}{5} \tan ^{5} t+C ($since $u = \tan t)$
$= \frac{2}{5} \tan ^{5} \sqrt{x}+C$ (since $t=\sqrt{x}$)
Hence, $\int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} d x=\frac{2}{5} \tan ^{5} \sqrt{x}+C$

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