Question

integrate (x^2)/(cos^2 x)

Answer 1

\[\int\limits \frac{ x^2 }{ \cos^2(x) } dx\] is this correct?

Answer 2

Yup..
should i factorise the square 1st? or using substitution?

Answer 3

Hm, this one is tricky. Have you tried integrating by parts? Changing cos to sec?

Answer 4

U mean integrate \[\int\limits_{}^{} x ^{2} \sec ^{2}x dx\]
then let
\[u = x ^{2}\]
\[v = \int\limits_{}^{} \sec ^{2}x\]

So,will hv
\[x ^{2}\tan x - \int\limits_{}^{} 2x \tan x dx \]

Answer 5

Yep. Use inegration by parts three times. It should work out.

Answer 6

I will help you out by telling you for the second integration\[v = \ln|\cos(x)| \]

Answer 7

I think it's best if you just memorize the integral of tan(x) because it's used often in integral calculus courses and physics courses.

Answer 8

So \[\int\limits_{}^{}2x \tan x dx = -2x \ln \cos x +2 \int\limits_{}^{} \ln \cos x dx\]
then,i have to integrate again for the lncosx,is it?

Answer 9

Ah. It all worked out in my head, but now that I see it written down, it doesn't look right. Haha. Um... have you learned how to integrate using complex exponentials? Haha. I actually am stump'd now. Haha. sorry for the confusion.

Answer 10

My second suggestion was going to be.
\[\int\limits x^2(1-\tan^2(x))dx\]

Answer 11

Haha. Its ok anyway.
The question actually sounds like this
\[\int\limits_{}^{} \frac{ x \sin 2x + x ^{2} }{ \cos ^{2}x } dx\]
and i separate it into two parts. so \[\int\limits_{}^{} \frac{ x ^{2} }{ \cos ^{2} x}\]
is the subpart of it.

Answer 12

Oh, um, well, this is just my train of thought, not sure if it's right though..
\[\int\limits \frac{ x(2\sin(x)\cos(x))+x^2 }{ \cos^2(x) }dx \]
Then you get:
\[\int\limits \frac{ 2xsin(x)+x^2 }{ \cos(x) }\]
Separate:
\[\int\limits \frac{ 2xsin(x) }{ \cos(x) }dx + \int\limits \frac{ x^2 }{ \cos(x) }dx\]
which is basically integration by parts for BOTH
\[2 \int\limits xtan(x)dx + \int\limits x^2 \sec(x)dx\]

Answer 13

It supposed to be \[\cos ^{2}x\] for the denominator

Answer 14

I understand, but one cos(x) get's cancel'd out for the first one. because you have:
\[\frac{ 2x \sin(x)\cos(x) }{ \cos^2(x) } = \frac{ 2xsin(x) }{ \cos(x) }\]