Integration using trigonometric identities

Some integrals look trigonometric but are not "standard" at all — ∫sin⁡2x dx\int \sin^2 x\,dx, ∫sin⁡3x dx\int \sin^3 x\,dx, ∫sin⁡3xcos⁡2x dx\int \sin 3x\cos 2x\,dx have no direct entry in the standard integral table. Integration using trigonometric identities is the fix: before reaching for any integration technique, rewrite the integrand using an identity you already know — double-angle, Pythagorean, or product-to-sum — so it collapses into a sum of standard terms or a substitution-ready form. No new calculus rule is introduced here; the entire skill is algebraic preparation. Get the identity right and the integral that follows is often routine. This article sits inside the Integrals chapter of CBSE Class 12 Mathematics and leans on identities from Class 11 trigonometry, then finishes with ordinary substitution or standard trig integrals.

The core insight

Rewrite a non-standard trigonometric integrand using a double-angle, Pythagorean, or product-to-sum identity until it becomes a sum of standard integrals or a substitution-ready expression, then integrate term by term.

The core idea

Trigonometric integrands rarely arrive in integrable form. The job before any integration happens is to recognize which identity converts the given expression into something the standard integral table (or a simple substitution) can handle.

Even powers — power reduction. A power like sin⁡2x\sin^2 x or cos⁡2x\cos^2 x has no direct antiderivative rule, but the double-angle identities rewrite it as a linear combination of 11 and cos⁡2x\cos 2x, both standard:

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2\sin^2 x = \frac{1-\cos 2x}{2}, \qquad \cos^2 x = \frac{1+\cos 2x}{2}

where: xx is the variable of integration, and cos⁡2x\cos 2x is the double-angle cosine (from cos⁡2x=1−2sin⁡2x=2cos⁡2x−1\cos 2x = 1-2\sin^2x = 2\cos^2x-1). Once expanded, each term integrates on its own.

Odd powers — peel and substitute. For an odd power such as sin⁡3x\sin^3 x, peel off a single factor and convert the remaining even power with a Pythagorean identity, which sets up a clean substitution:

sin⁡2x+cos⁡2x=1,1+tan⁡2x=sec⁡2x,1+cot⁡2x=csc⁡2x\sin^2 x + \cos^2 x = 1, \qquad 1+\tan^2x=\sec^2x, \qquad 1+\cot^2x=\csc^2x

where: these three identities let you trade sin⁡2x\sin^2x for 1−cos⁡2x1-\cos^2x (or the mirror swap for cosine), or replace tan⁡2x\tan^2x and cot⁡2x\cot^2x with expressions in sec⁡2x\sec^2x or csc⁡2x\csc^2x that integrate directly.

Products — product-to-sum. An integrand like sin⁡3xcos⁡2x\sin 3x\cos 2x is a product of two trig functions of different angles, which is never standard on its own. The product-to-sum formulas turn any such product into a sum of sines or cosines of single angles:

sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \tfrac12\big[\sin(A+B)+\sin(A-B)\big]
cos⁡Acos⁡B=12[cos⁡(A+B)+cos⁡(A−B)],sin⁡Asin⁡B=12[cos⁡(A−B)−cos⁡(A+B)]\cos A\cos B = \tfrac12\big[\cos(A+B)+\cos(A-B)\big], \qquad \sin A\sin B = \tfrac12\big[\cos(A-B)-\cos(A+B)\big]

where: AA and BB are the two angles in the product (e.g. A=3xA=3x, B=2xB=2x). Each resulting term is a standard ∫cos⁡(kx) dx\int \cos(kx)\,dx or ∫sin⁡(kx) dx\int \sin(kx)\,dx integral. The three routes share one shape: spot the non-standard piece, name the identity that dissolves it, rewrite, then integrate term by term. Higher even powers such as sin⁡4x\sin^4x need the power-reduction identity applied twice — first to sin⁡2x\sin^2x, then again to the resulting cos⁡22x\cos^2 2x term — since "reduce, expand, reduce again" always terminates.

Figure
Three trigonometric integrands routed through the matching identity into standard integrable terms
Match the pattern to the identity, then integrate even power: sin^2 x double-angle identity (1 - cos 2x) / 2 standard terms, integrate odd power: sin^3 x Pythagorean identity (1 - cos^2 x) sin x let t = cos x, substitute product: sin 3x cos 2x product-to-sum identity half sin 5x + half sin x two standard sine integrals standard integral table
Derivation AReducing sin to the fourth power to standard integrals

Establish, using only the double-angle identity applied twice and term-by-term integration, that ∫sin⁡4x dx=3x8−sin⁡2x4+sin⁡4x32+C\int \sin^4 x\,dx = \dfrac{3x}{8} - \dfrac{\sin 2x}{4} + \dfrac{\sin 4x}{32} + C, so that any even power beyond sin⁡2x\sin^2x reduces the same way — repeat power reduction until every term is a standard cosine integral, no matter how high the even power starts.

Step 1. Write sin⁡4x\sin^4x as (sin⁡2x)2(\sin^2x)^2 and apply the power-reduction identity once:

sin⁡4x=(1−cos⁡2x2)2=1−2cos⁡2x+cos⁡22x4\sin^4 x = \left(\frac{1-\cos 2x}{2}\right)^2 = \frac{1 - 2\cos 2x + \cos^2 2x}{4}

Step 2. The middle term cos⁡2x\cos 2x is already standard, but cos⁡22x\cos^2 2x is a squared cosine again — apply the same power-reduction identity, now with angle 2x2x in place of xx: cos⁡22x=1+cos⁡4x2\cos^2 2x = \dfrac{1+\cos 4x}{2}. Substituting this back:

sin⁡4x=14[1−2cos⁡2x+1+cos⁡4x2]=38−cos⁡2x2+cos⁡4x8\sin^4x = \frac{1}{4}\left[1 - 2\cos 2x + \frac{1+\cos 4x}{2}\right] = \frac{3}{8} - \frac{\cos 2x}{2} + \frac{\cos 4x}{8}

Step 3. Every term is now a standard integral. Integrate term by term using ∫cos⁡(kx) dx=sin⁡(kx)k+C\int \cos(kx)\,dx = \dfrac{\sin(kx)}{k}+C:

∫sin⁡4x dx=38∫dx−12∫cos⁡2x dx+18∫cos⁡4x dx=3x8−12⋅sin⁡2x2+18⋅sin⁡4x4+C\int \sin^4x\,dx = \frac{3}{8}\int dx - \frac12\int \cos 2x\,dx + \frac18\int \cos 4x\,dx = \frac{3x}{8} - \frac12\cdot\frac{\sin 2x}{2} + \frac18\cdot\frac{\sin 4x}{4} + C

Step 4. Simplify the coefficients to reach the final closed form:

∫sin⁡4x dx=3x8−sin⁡2x4+sin⁡4x32+C\int \sin^4x\,dx = \frac{3x}{8} - \frac{\sin 2x}{4} + \frac{\sin 4x}{32} + C

This confirms the "reduce, expand, reduce again" pattern from the core idea: a fourth power needed the identity twice, and the process always terminates because each reduction strictly lowers the trigonometric power inside the squared term.

Worked example

Evaluate ∫sin⁡3xcos⁡2x dx\displaystyle\int \sin 3x\cos 2x\,dx.

Step 1. The integrand is a product of sine and cosine at two different angles (A=3xA=3x, B=2xB=2x), so it is not standard. Apply the product-to-sum identity sin⁡Acos⁡B=12[sin⁡(A+B)+sin⁡(A−B)]\sin A\cos B = \tfrac12[\sin(A+B)+\sin(A-B)]:

sin⁡3xcos⁡2x=12[sin⁡(3x+2x)+sin⁡(3x−2x)]=12[sin⁡5x+sin⁡x]\sin 3x\cos 2x = \frac12\big[\sin(3x+2x)+\sin(3x-2x)\big] = \frac12\big[\sin 5x + \sin x\big]

Step 2. Both terms are now standard sine integrals. Integrate term by term using ∫sin⁡(kx) dx=−cos⁡(kx)k+C\int \sin(kx)\,dx = -\dfrac{\cos(kx)}{k}+C:

∫sin⁡3xcos⁡2x dx=12∫sin⁡5x dx+12∫sin⁡x dx=12(−cos⁡5x5)+12(−cos⁡x)+C\int \sin 3x\cos 2x\,dx = \frac12\int \sin 5x\,dx + \frac12\int \sin x\,dx = \frac12\left(-\frac{\cos 5x}{5}\right) + \frac12(-\cos x) + C

Step 3. Simplify:

∫sin⁡3xcos⁡2x dx=−cos⁡5x10−cos⁡x2+C\int \sin 3x\cos 2x\,dx = -\frac{\cos 5x}{10} - \frac{\cos x}{2} + C

Final answer: ∫sin⁡3xcos⁡2x dx=−cos⁡5x10−cos⁡x2+C\displaystyle\int \sin 3x\cos 2x\,dx = -\dfrac{\cos 5x}{10} - \dfrac{\cos x}{2} + C.

Competitive stretch

For ∫tan⁡2x dx\int \tan^2 x\,dx, no product-to-sum or power-reduction is needed — the Pythagorean identity tan⁡2x=sec⁡2x−1\tan^2x = \sec^2x - 1 finishes the problem outright: ∫tan⁡2x dx=∫(sec⁡2x−1) dx=tan⁡x−x+C\int \tan^2x\,dx = \int(\sec^2x-1)\,dx = \tan x - x + C. Recognizing when a single identity swap ends the problem, versus when it merely sets up a further substitution, is itself an exam skill.

Common traps

  • Trying to integrate sin⁡2x\sin^2x or cos⁡2x\cos^2x directly as if a power rule applied — these are not standard forms; power-reduction must come first.
  • Using the even-power method (power reduction) on an odd power, or the odd-power method (peel and substitute) on an even power — the two routes are not interchangeable.
  • Sign slips when swapping cos⁡2x=1−sin⁡2x\cos^2x = 1-\sin^2x or sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x — writing sin⁡2x−1\sin^2x-1 instead of 1−sin⁡2x1-\sin^2x flips the whole answer.
  • Mixing up the sign pattern across the three product-to-sum formulas, or forgetting the chain-rule factor kk in ∫cos⁡(kx) dx=sin⁡(kx)/k+C\int \cos(kx)\,dx = \sin(kx)/k + C.
  • Stopping after one power-reduction step on sin⁡4x\sin^4x or cos⁡4x\cos^4x and leaving a squared cosine term unintegrated — it needs a second reduction.

Ask the tutor: describe your trigonometric integrand to the AI tutor and ask which identity — power-reduction, peel-and-substitute, or product-to-sum — should be applied first, then request a few more practice integrands mixing even powers, odd powers, and products.

How do I know whether to use power-reduction or the peel-and-substitute method?

Look at whether the power is even or odd. Even powers like sin⁡2x\sin^2x, cos⁡4x\cos^4x have no factor left over to absorb a dtdt, so power-reduction (turning them into a sum of 11, cos⁡2x\cos 2x, cos⁡4x\cos 4x, etc.) is the only route. Odd powers like sin⁡3x\sin^3x, cos⁡5x\cos^5x always have one leftover factor after pairing up the rest into an even power, and that leftover factor becomes exactly the dtdt a substitution needs. So: even power means power-reduction, odd power means peel one factor and substitute.

Do I need to memorize the expanded sin^4 x and cos^4 x formulas?

No — memorize the method, not the expanded result. The double-angle identity sin⁡2x=(1−cos⁡2x)/2\sin^2x = (1-\cos 2x)/2 (and its cosine twin) is the only thing you need to recall; apply it once for a square, and apply it a second time to whatever squared cosine term appears inside the first expansion. Re-deriving sin⁡4x\sin^4x in an exam takes under a minute once the pattern is automatic, and it protects you from misremembering a coefficient.

Why does the product-to-sum formula work — is it just an algebraic trick?

It follows directly from the angle-addition formulas you already know: sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B and sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B)=\sin A\cos B-\cos A\sin B. Adding these two equations cancels the cos⁡Asin⁡B\cos A\sin B terms and leaves sin⁡(A+B)+sin⁡(A−B)=2sin⁡Acos⁡B\sin(A+B)+\sin(A-B)=2\sin A\cos B, which rearranges to the product-to-sum formula used above. So it is not a separate rule to trust blindly — it is a direct consequence of angle addition, and you can re-derive it in seconds if you forget the exact form.

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