Integration with trig functions

• November 24th 2009, 04:04 AM
Beard
Integration with trig functions
Hi I'm stuck on a problem (I don't know if I am doing it correctly):

$\int \frac{1}{1 + x^2}.dx$
I have been given a hint to substitute x^2 with tan^2(u).

When I do this it becomes $\frac{1}{1 + tan^2(u)}.dx$ now using the identity of tan^2 = sec^2 - 1 i get $\frac{1}{\sec^2(u)}.dx$ it becomes $\cos^2(u).dx$.

I know that to make it so the integral is with respect to (u) I find the differential of x= tan^2(u) which when I rearrange becomes $dx = sec^2(u).du therefore \int cos^2(u)\sec^2(u).du$. Here is where I am confused, does the above just become $\int 1$ because sec and cos cancelling each other out? or is there something that I have missed out?

Thanks for any help
• November 24th 2009, 04:08 AM
mr fantastic
Quote:

Originally Posted by Beard
Hi I'm stuck on a problem (I don't know if I am doing it correctly):

$\int \frac{1}{1 + x^2}.dx$
I have been given a hint to substitute x^2 with tan^2(u).

When I do this it becomes $\frac{1}{1 + tan^2(u)}.dx$ now using the identity of tan^2 = sec^2 - 1 i get $\frac{1}{\sec^2(u)}.dx$ it becomes $\cos^2(u).dx$.

I know that to make it so the integral is with respect to (u) I find the differential of x= tan^2(u) which when I rearrange becomes $dx = sec^2(u).du therefore \int cos^2(u)\sec^2(u).du$. Here is where I am confused, does the above just become $\int 1$ Mr F says: No. It becomes ${\color{red}\int 1 du = u + C}$ ....

because sec and cos cancelling each other out? or is there something that I have missed out?

Thanks for any help

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