# Evaluate the integral. \tan x\sec^{3}xdx

Evaluate the integral.
$$\displaystyle{\tan{{x}}}{{\sec}^{{{3}}}{x}}{\left.{d}{x}\right.}$$

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Alex Sheppard
Step 1
Given
Evaluate the integral.
$$\displaystyle\int{\tan{{x}}}{{\sec}^{{{3}}}{x}}{\left.{d}{x}\right.}$$
Step 2
Let $$\displaystyle{u}={\sec{{\left({x}\right)}}}.$$
Then
$$\displaystyle{d}{u}={\left({\sec{{\left({x}\right)}}}\right)}'{\left.{d}{x}\right.}$$
$$\displaystyle={\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}{d}{u}$$
$$\displaystyle={\left({\sec{{\left({x}\right)}}}\right)}'{\left.{d}{x}\right.}$$
$$\displaystyle={\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
,and we have that
$$\displaystyle{\tan{{\left({x}\right)}}}{\sec{{\left({x}\right)}}}{\left.{d}{x}\right.}={d}{u}.$$
Therefore,
$$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}^{{{3}}}{\left({x}\right)}}{\left.{d}{x}\right.}=\int{u}^{{{2}}}{d}{u}$$
Apply the power rule $$\displaystyle\int{u}^{{{n}}}{d}{u}={\frac{{{u}^{{{n}+{1}}}}}{{{n}+{1}}}}{\left({n}\ne-{1}\right)}$$
$$\displaystyle\int{u}^{{{2}}}{d}{u}={\frac{{{u}^{{{1}+{2}}}}}{{{1}+{2}}}}={\left({\frac{{{u}^{{{3}}}}}{{{3}}}}\right)}$$
Recall that $$\displaystyle{u}={\sec{{\left({x}\right)}}}{u}={\sec{{\left({x}\right)}}}:$$
$$\displaystyle{\frac{{{u}^{{{3}}}}}{{{3}}}}={\frac{{{\sec{{\left({x}\right)}}}^{{{3}}}}}{{{3}}}}$$
Therefore,
$$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}^{{{3}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\frac{{{{\sec}^{{{3}}}{\left({x}\right)}}}}{{{3}}}}$$
$$\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}^{{{3}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\frac{{{{\sec}^{{{3}}}{\left({x}\right)}}}}{{{3}}}}+{C}$$
###### Not exactly what you’re looking for?
$\int \tan x\sec^{3}xdx=\int \sec^{2}x\tan x\sec x dx=\begin{bmatrix}u=\sec x \\du=\tan x\sec xdx \end{bmatrix}$
$$\displaystyle=\int{u}^{{{2}}}{d}{u}={\frac{{{u}^{{{3}}}}}{{{3}}}}+{C}={\frac{{{{\sec}^{{{3}}}{x}}}}{{{3}}}}+{C}$$
Result:
$$\displaystyle{\frac{{{{\sec}^{{{3}}}{x}}}}{{{3}}}}+{C}$$
karton

$$\\\text{Step 1} \\\text{Apply u-substitution} \\\text{Let}\ u=\sec^{3}(x)\Rightarrow dx=\frac{1}{3\sec^{3}(x)\tan(x)} \\\text{Thus,} \\\int \tan(x)\sec^{3}(x)dx=\frac{1}{3}\int 1du \\=\frac{1}{3}u+C \\=\frac{\sec^{3}(x)}{3}+C \\\text{Result:} \\\frac{\sec^{3}(x)}{3}+C$$