Question

Evaluate the following integral:
$\int{\tan x{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}dx}$

Answer 1

Hint: To solve this question substitute value of $1-{{\tan }^{2}}x=t$
We have the given integral as $I=\int{\tan x{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}dx}..........\left( 1 \right)$

Here, we can use substitute method for finding/solving the given integral in a proper way:
Let $t=1-{{\tan }^{2}}x$
Differentiating both sides with respect to $x$
$t=1-{{\tan }^{2}}x$
$\dfrac{dt}{dx}=-2\tan x{{\sec }^{2}}x$ $\left( \dfrac{d}{dx}\left( \tan x \right)\ And -{{\sec }^{2}}x \right)$ chain rule is applied
$dt=-2\tan x{{\sec }^{2}}xdx.............\left( 2 \right)$
From the equation $\left( 1 \right)\And \left( 2 \right)$; we can replace $\tan x{{\sec }^{2}}xdx$ by
above equation $\left( 2 \right)$ as
$\tan x{{\sec }^{2}}xdx=\dfrac{-dt}{2}$
Hence, equation $\left( 1 \right)$ will become
$I=\int{\dfrac{-1}{2}\sqrt{t}dt}$ as $\left( 1-{{\tan }^{2}}x=t \right)$

& I=\dfrac{-1}{2}\int{{{t}^{\dfrac{1}{2}}}}dt \\\ & I=\dfrac{-1}{2}\dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1}+C\text{ } as \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\\ \end{aligned}$$ $$I=\dfrac{-1}{2}\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+C=\dfrac{-1}{2}\times \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}+C$$ $$\begin{aligned} & I=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}+C \\\ & \text{As }t=1-{{\tan }^{2}}x \\\ & I=\dfrac{-1}{3}{{\left( 1-{{\tan }^{2}}x \right)}^{\dfrac{3}{2}}}+C \\\ \end{aligned}$$ Hence, $\int{\tan x{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}}dx=\dfrac{-1}{3}{{\left( 1-{{\tan }^{2}}x \right)}^{\dfrac{3}{2}}}+C$ Note: One can substitute $t=\tan x$ Hence $dt={{\sec }^{2}}xdx$ and then can put value in integral. Therefore $I=\int{t\sqrt{1-{{t}^{2}}}dt}$ Now, he/she needs to put ${{t}^{2}}=y\And 1-{{t}^{2}}=y$ to solve the above integral. Hence, it takes one more step than the solution provided but the answer will be the same. One can convert $\tan x\And {{\sec }^{2}}x$ to cosine and sine forms which students do generally will take more time as well.