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

Last updated date: 15th Jul 2024
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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}} \\
& 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 \\
$\int{\tan x{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}}dx=\dfrac{-1}{3}{{\left( 1-{{\tan }^{2}}x

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.