Question

Integral Problem

Answer 1

@ganeshie8

Answer 2

$$\int_{0}^{\frac{\pi}{2}} \frac{sin(2n+1)x}{sinx} dx = \frac{\pi}{2}$$ prove by induction that it is true for every n>o and n=0

Answer 3

I think you'll want to expand the numerator out. Try that.

Answer 4

$$\int_{0}^{\pi/2} sin(2nx)cotx dx$$ . I am correct till here ??

Answer 5

you're missing one more term, a cos(2nx)

Answer 6

I found the other part to be = 0

Answer 7

ok. Here's what you will need to do.
Prove the statement is true for n = 1. That is not hard.
Assume it is true for n = k.

then write out the expansion for n = k+1.
\[\frac{ \sin((2k+1)+1)x }{ sinx } = \frac{ \sin(2k+1)xcos(2x)+\sin(2x)\cos(2k+1)x }{ sinx }\]

Simplify the sin2x and cos2x in terms of sinx and you'll be not far off from your answer. remember to include the assumption

Answer 8

I found this method in MSE

Answer 9

Well from this step: sin((2k+1)+1)xsinx=sin(2k+1)xcos(2x)+sin(2x)cos(2k+1)xsinx
expand cos(2x) as 2sin^2(x) - 1 and sin2x as 2sinxcosx and dividing by sin x on the denominator will generate 2sinx sin(2(k+1)x + 2cos(2(k+1)xcosx (which can be simplified into a cosine expansion!) as well as your parent integral.

Answer 10

Oops I made a bit of a mistake in my writing which might have confused you. For the n = k+1 step:

\[\sin(2(k+1)+1)x = \sin((2(k+1)x + 2x) = \sin(2(k+1))xcos(2x) + \cos(2(k+1)x)\sin(2x)\]

Answer 11

That's using the sine expansion for sin(A+B) where A = 2(k+1)x and B = 2x
Then
\[\sin(2(k+1))xcos(2x) = \sin(2(k+1)x(1-2\sin^2(x)) \]
and \[\cos(2(k+1)x)\sin(2x) = 2\sin(x)\cos(x)\cos(2(k+1))x\]

Divide these by sinx and a special result will show

Answer 12

Thank you very much. :)