Question

compute this definite integral using method of substitution

Answer 1

\[\int\limits_{0}^{\frac{ \Pi }{ 4 }}\tan^2(x)+\tan^4(x) dx\]

Answer 2

I'm not sure what I should make "u"

Answer 3

I know that tan^2x= sec^2x-1, but I'm confused about how to set up the problem

Answer 4

factor out a \(\tan^2(x)\)

Answer 5

\[\int\limits_{?}^{?}\tan^2(x)(1-\tan^2(x))\] like that?

Answer 6

It should be +

Answer 7

whoops, yeah. What do I next then?

Answer 8

yeah, everyone seems to run away from this problem......

Answer 9

If you set u = tan x, then du (1+tan²x)dx, so \[\int\limits_{}^{}\tan^2x(1+\tan^2x)dx=\int\limits_{}^{}u^2du\]

Answer 10

(typo): du=(1+tan²x)dx

Answer 11

I'm sorry, I'm a little confused still. isn't the derivative of tan(x) sec^2x?

Answer 12

You can write (tan x)' in several ways:\[(\tan x)'=\left( \frac{ \sin x }{ \cos x } \right)'=\frac{ \cos x \cdot \cos x-sinx \cdot -\sin x }{ \cos^2x }=\frac{ \cos^2x+\sin^2x }{ \cos^2 x }\]From here you have several options. Because sin²x+cos²x=1, it can be written as:\[\frac{ 1 }{ \cos^2x }=\sec^2x\]YOu can also split up the fraction:\[\frac{ \cos^2x }{ \cos^2x }+\frac{ \sin^2x }{ \cos^2x }=1+\tan^2x\]