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Integrals like ∫ sin³x cos²x dx or ∫ sin⁴x dx come up in JEE on their own and in the middle of longer problems, such as areas and definite integrals. They are not hard once you know the one decision that matters: is either power odd? This post explains the two cases, works four examples, and shows how to check an answer that looks different from the key.
What does the syllabus cover?
The JEE (Main) 2026 syllabus lists "Integration using trigonometric identities" in Unit 8. Section 7.3.2 of the current NCERT Class 12 chapter on integrals covers it. Its Example 7 finds ∫ cos²x dx with cos 2x = 2cos²x − 1, and ∫ sin³x dx in two ways. Check the syllabus for your exam year.
Why does an odd power make it easy?
Because an odd power lets you set one factor aside as the derivative of the substitution.
Say sin x has an odd power. Write sin³x = sin²x · sin x = (1 − cos²x) sin x. Now everything except the single sin x is in terms of cos x, and sin x dx is exactly −d(cos x). Put t = cos x and the integral becomes a polynomial in t.
If cos x has the odd power, swap the roles: keep one cos x, convert the rest with cos²x = 1 − sin²x, and put t = sin x.
Worked example 1: sin³x cos²x
The power of sin is odd. Keep one sin x:
∫ sin³x cos²x dx = ∫ (1 − cos²x) cos²x sin x dx.
Put t = cos x, dt = −sin x dx:
= −∫ (t² − t⁴) dt = −t³/3 + t⁵/5 + C = −cos³x/3 + cos⁵x/5 + C.
Worked example 2: cos⁵x
The power of cos is odd. Keep one cos x and write cos⁴x = (1 − sin²x)²:
∫ cos⁵x dx = ∫ (1 − sin²x)² cos x dx.
Put t = sin x, dt = cos x dx:
= ∫ (1 − 2t² + t⁴) dt = sin x − (2/3)sin³x + (1/5)sin⁵x + C.
What do you do when both powers are even?
Then there is no spare factor, and substitution does not work. Instead, lower the powers with the half-angle identities:
sin²x = (1 − cos 2x)/2
cos²x = (1 + cos 2x)/2
sin x cos x = (sin 2x)/2
Apply them until every term is a single sine or cosine of a multiple of x.
Worked example 3: sin⁴x
sin⁴x = [(1 − cos 2x)/2]² = (1 − 2cos 2x + cos²2x)/4.
The cos²2x is still a square, so use the identity again: cos²2x = (1 + cos 4x)/2. Then
sin⁴x = 3/8 − (1/2)cos 2x + (1/8)cos 4x,
and ∫ sin⁴x dx = 3x/8 − (sin 2x)/4 + (sin 4x)/32 + C.
As a definite integral from 0 to π/2, both sine terms vanish at the limits, so the value is (3/8)(π/2) = 3π/16.
Worked example 4: sin²x cos²x
Both powers are even. Use sin x cos x = (sin 2x)/2 first:
sin²x cos²x = (sin²2x)/4 = (1 − cos 4x)/8.
So ∫ sin²x cos²x dx = x/8 − (sin 4x)/32 + C.
Which mistakes cost the most marks?
Using half-angle formulas when one power is odd. It works, but takes three times as long.
Losing the minus sign from dt = −sin x dx.
Stopping after one round of half-angle when a cos² term remains, as in example 3.
Writing ∫ cos 4x dx = sin 4x instead of (sin 4x)/4.
Marking a correct answer wrong because it uses a different identity.
Why practise choosing, not just solving?
Sweller and Cooper (1985) found that studying worked examples helped students solve later problems faster and with fewer errors, but only problems with the same structure as the examples. Odd and even cases look alike on the page, so the skill to train is the first glance: which power is odd? Mixed practice trains exactly that. In a study of 126 seventh-grade students, interleaved practice produced higher test scores than blocked practice one day and 30 days later, because it "requires students to choose a strategy on the basis of the problem itself".
A practice routine
Sort first (10 minutes, three times a week). For ten integrals of sinᵐx cosⁿx, write only "odd: t = cos x", "odd: t = sin x" or "even: half-angle".
Solve five, then differentiate each answer to check it.
One definite integral a day from 0 to π/2, which you can also check against the known value of a related integral.
Mix in other techniques from our integration practice plan.
If a reduction keeps going wrong and you cannot see where, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each line.
Today, find ∫ sin²x cos³x dx by the odd-power method, then check it by differentiating.
