Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Integrating trig quick question!?

For this problem:

\int \frac{4}{sin^2x}\ dx

I have gone about my working in this way:

\int \frac{4}{sin^2x}\ dx = -1 \times 4 \times -cosx \times (sinx)^{-1} +C

=

\frac{4cosx}{sinx}\ + C

I think I have done something wrong.... :s

(edited 8 years ago)

if you integrate sec² you get tan

if you integrate cosec² you get.....

OP

Original post by the bear

if you integrate sec² you get tan

if you integrate cosec² you get.....

Wait.. why is this to do with sec² and cosec²'s ?

Original post by Funky_Giraffe

Wait.. why is this to do with sec² and cosec²'s ?

because

\frac{1}{sin^2x}\

is cosec²x

imsoanonymous123

It's not exactly clear what you did in your workings but you appear to be missing a minus sign

WhiteGroupMaths

Original post by Funky_Giraffe

For this problem:

\int \frac{4}{sin^2x}\ dx

I have gone about my working in this way:

\int \frac{4}{sin^2x}\ dx = -1 \times 4 \times -cosx \times (sinx)^{-1} +C

=

\frac{4cosx}{sinx}\ + C

I think I have done something wrong.... :s

Your workings are incorrect.

Inspecting what you have written, I believe you were attempting to use the proven result of

\int [f(x)]^n * f'(x) dx

=

\frac{ [f(x)]^{n+1}} {(n+1)}\ +C

, where f(x)= sin x and n= -2.

Unfortunately, this is not feasible because f'(x)=cos x is not present in the original integral. For some unknown reason you made this term surface all of a sudden, which makes no sense.

Like the rest have mentioned, there is a simple result for integrating cosec^2 x wrt x.

Hope this helps. Peace.

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