Q.Integrate the following function: sin3xcos4x
Concept understanding — Product To Sum Identity
Why turn a product into a sum?
Products of trig functions are messy — they don't integrate nicely and are hard to simplify. Sums are clean: you can separate and integrate them term-by-term. The Product-to-Sum identities convert a product of sines and cosines into a sum (or difference), turning something like sin3xcos5x into 21[sin8x+sin(−2x)].
The core idea in one sentence
Any product of two sines and/or cosines can be rewritten as half the sum (or difference) of two sine/cosine functions whose arguments are the sum and difference of the original angles.
The four identities
For any two angles A and B:
sinAcosBcosAsinBcosAcosBsinAsinB=21[sin(A+B)+sin(A−B)]=21[sin(A+B)−sin(A−B)]=21[cos(A+B)+cos(A−B)]=21[cos(A−B)−cos(A+B)]
The pattern:
- Same functions (coscos or sinsin) → result uses cosines.
- Different functions (sincos or cossin) → result uses sines.
- The sinsin case has a minus before cos(A+B) — the one that trips people up.
Where they come from (the derivation)
They follow directly from the sum and difference formulas:
sin(A+B)sin(A−B)cos(A+B)cos(A−B)=sinAcosB+cosAsinB=sinAcosB−cosAsinB=cosAcosB−sinAsinB=cosAcosB+sinAsinB
Add the first two: sin(A+B)+sin(A−B)=2sinAcosB. Divide by 2 → the first identity. Subtract them → the second. Add/subtract the cosine formulas → the other two.
If you forget an identity, derive it in 10 seconds from the sum/difference formulas.
Worked examples
Simplify sin5xcos2x. With A=5x, B=2x (first identity):
sin5xcos2x=21[sin7x+sin3x]
Simplify sin4xsinx. With A=4x, B=x (fourth identity):
sin4xsinx=21[cos3x−cos5x]
The sinsin identity has a minus between the two cosines, with cos(A−B) first. Writing 21[cos(A+B)−cos(A−B)] is wrong.
When will you use this?
- Integration: ∫sin3xcos5xdx becomes 21∫(sin8x+sin(−2x))dx — trivial.
- Solving equations and physics (wave interference, signal processing), where products of sinusoids appear constantly.
Doubt yourself? Test with a simple angle. With A=30∘, B=0∘: sin30∘cos0∘=0.5, and 21[sin30∘+sin30∘]=0.5. ✓
Bottom line: Product-to-sum identities turn multiplication into addition — and addition is always easier to handle.
Product-to-sum identities are part of the NCERT Class 11 Trigonometric Functions chapter and become essential again in the Class 12 Integrals chapter whenever a product like sin 3x cos 5x needs to be integrated. Students searching 'product to sum formulas class 11 trigonometry' or 'how to integrate sin x cos x product' will find these four identities are exactly the transformation tool both CBSE units expect students to have memorized.
Turn the product into a sum with sinAcosB=21[sin(A+B)+sin(A−B)].
With A=3x, B=4x:
sin3xcos4x=21[sin7x+sin(−x)]=21[sin7x−sinx].
Integrate term by term (using ∫sinkxdx=−k1coskx):
∫sin3xcos4xdx=21(−7cos7x+cosx)+C=−141cos7x+21cosx+C.
∫sin3xcos4xdx=−141cos7x+21cosx+C
Product-to-sum gives sin3xcos4x=21(sin7x−sinx), and integrating gives −141cos7x+21cosx+C.
Why convert to a sum
Products of sines and cosines are hard to integrate directly, but sums are trivial. The identity sinAcosB=21[sin(A+B)+sin(A−B)] does the conversion.
Apply the identity
Take A=3x, B=4x:
sin3xcos4x=21[sin(7x)+sin(−x)].
Since sin(−x)=−sinx,
sin3xcos4x=21[sin7x−sinx].
Integrate
Use ∫sinkxdx=−k1coskx:
∫sin3xcos4xdx=21(−7cos7x)−21(−cosx)+C=−141cos7x+21cosx+C.
Watch the 71: 21⋅71=141, not 21.
∫sin3xcos4xdx=−141cos7x+21cosx+C
Method: Product-to-sum identity for sin(ax)cos(bx)
A product of a sine and a cosine of different angles cannot be integrated as-is; convert it into a sum of sines, which integrate term by term.
Steps
Step 1: Apply the correct product-to-sum identity.
sinAcosB=21[sin(A+B)+sin(A−B)]
Match A and B to the two angles in the integrand.
Step 2: Simplify signed angles using parity.
A negative angle inside sine flips sign: sin(−x)=−sinx. (For cosine, cos(−x)=cosx.) Simplify before integrating.
Step 3: Integrate each sine term.
Use ∫sin(kx)dx=−k1cos(kx)+C, keeping the k1 factor for every term.
Remember which identity to reach for: a sincos or cossin product yields sines, while sinsin or coscos yields cosines.
Common Mistakes
Mistake 1: Trying to integrate sin3xcos4x as a single product.
Why it's wrong: there is no u whose derivative appears because the two angles differ, so direct substitution fails. Correct approach: use sinAcosB=21[sin(A+B)+sin(A−B)] to split it into 21[sin7x+sin(−x)].
Mistake 2: Leaving sin(−x) without simplifying its sign.
Why it's wrong: sin(−x)=−sinx, so overlooking the odd symmetry gives a wrong sign on that term. Correct approach: simplify to 21[sin7x−sinx] before integrating, yielding −141cos7x+21cosx+C.
- KEAM 2025Set eng-2025-04294 marksMCQQ.The value of sin125πsin12π is equal to (A) 1 (B) 41 (C) 21 (D) 23 (E) 0
›Reveal solutionSolution
sin125πsin12π=sin75∘sin15∘=21[cos60∘−cos90∘]=21⋅21=41.
Using sinAsinB=21[cos(A−B)−cos(A+B)] with A=75∘,B=15∘: 21[cos60∘−cos90∘]=21[21−0]=41.
✓Final answerThe correct option is (B).
- KEAM 2024Set eng-2024-06054 marksMCQQ.cos18∘cos42∘cos78∘= (A) 41cos36∘ (B) 41cos72∘ (C) 41sin72∘ (D) 41sin36∘ (E) None of the above
›Reveal solutionSolution
Pairing cos42∘cos78∘ and combining reduces the product to 41sin36∘.
cos42∘cos78∘=21(cos120∘+cos36∘)=21(−21+cos36∘).
Multiplying by cos18∘ and evaluating numerically:
cos18∘cos42∘cos78∘≈(0.9511)(0.7431)(0.2079)≈0.1470.
And 41sin36∘=41(0.5878)=0.1470. ✓
✓Final answerThe correct option is (D).
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.Let k be a real number such that sin143πcos143π=kcos14π. Then the value of 4k is (A) 1 (B) 2 (C) 3 (D) 4 (E) 0
›Reveal solutionSolution
sin(3pi/14) cos(3pi/14) = (1/2) sin(3pi/7) = (1/2) cos(pi/14), so k = 1/2 and 4k = 2.
Concept and Intuition
The double-angle identity sin A cos A = (1/2) sin 2A, plus the co-function relation sin(pi/2 - x) = cos x, collapses the expression onto cos(pi/14).
Step-by-Step Solution
- sin(3pi/14) cos(3pi/14) = (1/2) sin(6pi/14) = (1/2) sin(3pi/7).
- 3pi/7 = pi/2 - pi/14, so sin(3pi/7) = cos(pi/14).
- Thus LHS = (1/2) cos(pi/14), giving k = 1/2.
- Therefore 4k = 2.
Common Mistakes
- Forgetting to convert sin(3pi/7) to cos(pi/14) via the complementary-angle identity.
✓Final answerThe correct option is (B) — 2.
ANSWER: B
- KEAM 2026Set eng-2026-04224 marksMCQQ.The value of 4cos36∘cos72∘ is equal to (A) 1 (B) 2 (C) 2 (D) 2 (E) 0
›Reveal solutionSolution
[!TLDR]
With cos36∘cos72∘=41, the expression 4cos36∘cos72∘=1.
Concept
The standard exact values are cos36∘=45+1 and cos72∘=45−1 (they can also be derived from multiple-angle identities in the NCERT/CBSE trigonometry syllabus).
Solution
cos36∘cos72∘=45+1⋅45−1=16(5)2−12=165−1=164=41.
Therefore
4cos36∘cos72∘=4×41=1.
[!ANSWER]
(A) 1
- KEAM 2025Set eng-2025-04234 marksMCQQ.Let f(x)=cosx. Then the value of 21[f(x+y)+f(y−x)]−f(x)f(y) is equal to (A) 2 (B) −2 (C) 1 (D) −1 (E) 0
›Reveal solutionSolution
The expression simplifies identically to 0.
Concept and Intuition
Using the sum-to-product identity cosP+cosQ, the bracket collapses to 2cos x cos y; halving it exactly cancels the product term cos x cos y.
Step-by-Step Solution
- cos(x+y)=cosxcosy−sinxsiny.
- cos(y−x)=cosycosx+sinysinx.
- Sum =2cosxcosy, so 21[⋯]=cosxcosy.
- Subtract f(x)f(y)=cosxcosy: result =0.
Common Mistakes
- Sign slips when expanding cos(y−x), which would prevent the sine terms from cancelling.
✓Final answerThe correct option is (E) — 0.
ANSWER: E
- KEAM 2024Set eng-2024-06054 marksMCQQ.cos7x+cos5xsin7x+sin5x= (A) sin6xtan6x (B) cos6xtan6x (C) sin6x (D) cos6x (E) tan6x
›Reveal solutionSolution
Apply sum-to-product to numerator and denominator; the cosx cancels leaving tan6x.
sin7x+sin5x=2sin6xcosx,cos7x+cos5x=2cos6xcosx.
2cos6xcosx2sin6xcosx=cos6xsin6x=tan6x.
✓Final answerThe correct option is (E).
- KEAM 2025Set eng-2025-04234 marksMCQQ.If sinx+siny=a, cosx+cosy=b and x+y=32π, then the value of ba is equal to (A) 33 (B) 23 (C) 3 (D) 43 (E) 63
›Reveal solutionSolution
a/b = tan((x+y)/2) = tan(pi/3) = sqrt3.
Concept and Intuition
Sum-to-product turns sin x + sin y and cos x + cos y into forms sharing the common factor 2 cos((x-y)/2), so their ratio is simply tan of the half-sum (x+y)/2.
Step-by-Step Solution
- a = sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2).
- b = cos x + cos y = 2 cos((x+y)/2) cos((x-y)/2).
- a/b = tan((x+y)/2) = tan(pi/3) = sqrt3.
Common Mistakes
- Trying to compute a and b separately instead of using the ratio, which needs no value of x - y.
✓Final answerThe correct option is (C) — sqrt3.
ANSWER: C
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