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

Concepcion Hale

Concepcion Hale

Answered question

2022-01-07

Evaluate the integral.
tanxsec3xdx

Answer & Explanation

Alex Sheppard

Alex Sheppard

Beginner2022-01-08Added 36 answers

Step 1
Given
Evaluate the integral.
tanxsec3xdx
Step 2
Let u=sec(x).
Then
du=(sec(x))dx
=tan(x)sec(x)dxdu
=(sec(x))dx
=tan(x)sec(x)dx
,and we have that
tan(x)sec(x)dx=du.
Therefore,
tan(x)sec3(x)dx=u2du
Apply the power rule undu=un+1n+1(n1)
u2du=u1+21+2=(u33)
Recall that u=sec(x)u=sec(x):
u33=sec(x)33
Therefore,
tan(x)sec3(x)dx=sec3(x)3
Add the constant of integration:
tan(x)sec3(x)dx=sec3(x)3+C
Pademagk71

Pademagk71

Beginner2022-01-09Added 34 answers

tanxsec3xdx=sec2xtanxsecxdx=[u=secxdu=tanxsecxdx]
=u2du=u33+C=sec3x3+C
Result:
sec3x3+C
karton

karton

Expert2022-01-11Added 613 answers

Step 1Apply u-substitutionLet u=sec3(x)dx=13sec3(x)tan(x)Thus,tan(x)sec3(x)dx=131du=13u+C=sec3(x)3+CResult:sec3(x)3+C

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