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

Concepcion Hale 2022-01-07 Answered
Evaluate the integral.
\(\displaystyle{\tan{{x}}}{{\sec}^{{{3}}}{x}}{\left.{d}{x}\right.}\)

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Expert Answer

Alex Sheppard
Answered 2022-01-08 Author has 1572 answers
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,
\(\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}^{{{3}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\frac{{{{\sec}^{{{3}}}{\left({x}\right)}}}}{{{3}}}}\)
Add the constant of integration:
\(\displaystyle\int{\tan{{\left({x}\right)}}}{{\sec}^{{{3}}}{\left({x}\right)}}{\left.{d}{x}\right.}={\frac{{{{\sec}^{{{3}}}{\left({x}\right)}}}}{{{3}}}}+{C}\)
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Pademagk71
Answered 2022-01-09 Author has 5116 answers
\[\int \tan x\sec^{3}xdx=\int \sec^{2}x\tan x\sec x dx=\begin{bmatrix}u=\sec x \\du=\tan x\sec xdx \end{bmatrix}\]
\(\displaystyle=\int{u}^{{{2}}}{d}{u}={\frac{{{u}^{{{3}}}}}{{{3}}}}+{C}={\frac{{{{\sec}^{{{3}}}{x}}}}{{{3}}}}+{C}\)
Result:
\(\displaystyle{\frac{{{{\sec}^{{{3}}}{x}}}}{{{3}}}}+{C}\)
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karton
Answered 2022-01-11 Author has 8659 answers

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

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