How do you compute the antiderivative sinxcos(2x)?

1 Answer
MusicallyEternal
Mar 6, 2015

The required integral is

I=∫sinxcos(2x)dx

The first part of the solution requires the use of the trigonometric identity cos(2x)=2cos2x−1

This leaves you with,

I=∫sinx2cos2x−1dx

Let cosx=t

Differentiating both sides with respect to t,

−sinxdxdt=1
⇒sinxdx=−dt

Thus,

I=−∫12t2−1dt

⇒I=−12∫1t2−12dt

This sort of integral has a standard result, which is

∫1x2−a2dx=12aln∣∣∣x−ax+a∣∣∣+C

which I will derive once we have the answer to your question, which is

I=−12⎡⎢ ⎢⎣12(1√2)ln∣∣ ∣∣t−1√2t+1√2∣∣ ∣∣⎤⎥ ⎥⎦+C

⇒I=−√24ln∣∣ ∣∣t−1√2t+1√2∣∣ ∣∣+C

Finally, replacing t=cosx, we get,

∫sinxcos(2x)dx=−√24ln∣∣ ∣∣cosx−1√2cosx+1√2∣∣ ∣∣+C

Now, for the derivation of ∫1x2−a2dx=12aln∣∣∣x−ax+a∣∣∣+C here's a screenshot from a textbook.

enter image source here

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