Question
Evaluate ∫sin2xdx using the double angle formula.
Solution — Step by Step
We cannot integrate sin2x directly because it’s not a standard form. The function sinx integrates to −cosx, but sin2x is a different beast.
The strategy: use a trigonometric identity to replace sin2x with an expression that we CAN integrate directly.
The relevant identity is:
cos2x=1−2sin2x
Solving for sin2x:
sin2x=21−cos2x
This converts a squared trig function into a simple cosine — which integrates easily.
∫sin2xdx=∫21−cos2xdx
=21∫(1−cos2x)dx
=21[∫1dx−∫cos2xdx]
=21[x−2sin2x]+C
∫sin2xdx=2x−4sin2x+C
We can also write this as 2x−41sin2x+C.
Why This Works
The double angle formula is the key to integrating even powers of sine and cosine. The identity cos2x=1−2sin2x comes from cos(A+B)=cosAcosB−sinAsinB with A=B=x, giving cos2x=cos2x−sin2x=1−2sin2x.
The technique generalises: for higher even powers like sin4x, we apply the identity twice. For odd powers like sin3x, a substitution approach works better.
The constant C in the answer is the constant of integration — never forget it in indefinite integrals.
Alternative Method
Using sin2x=21−cos2x, we can also write the answer as:
21(x−2sin2x)+C=21x−41sin2x+C
Both forms are equivalent. Some textbooks also write sin2x=2sinxcosx, giving 2x−2sinxcosx+C — verify this is the same using the double angle identity.
Common Mistake
The most common error is writing ∫sin2xdx=3sin3x+C — treating sin2x like a power function and using the power rule. This is WRONG. The power rule applies to polynomial powers of x, not to powers of sinx. For powers of trigonometric functions, always use trigonometric identities to reduce the power first.
The companion result ∫cos2xdx=2x+4sin2x+C follows from the identity cos2x=21+cos2x. Notice the only change is + instead of − inside the bracket. These two are among the most frequently appearing integrals in JEE — commit them to memory.