Q.Integrate: ∫sin2xcos2xdx
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Integration By Expansion
Integration by Expansion
The idea
Some integrands look forbidding only because they are written as a product or a power. If you first multiply them out — expand them — into a plain sum of standard terms, you can then integrate the sum term by term using the basic formulas you already know. Expansion is not a new rule of integration; it is a rewriting step that turns the integrand into something the sum rule and the power rule can finish.
This works because integration is linear: ∫(f±g)dx=∫fdx±∫gdx. Once the integrand is a sum, each piece is handled separately.
Algebraic expansion
Products and powers of polynomials are expanded first:
∫(x+2)2dx=∫(x2+4x+4)dx=3x3+2x2+4x+C.
Similarly, split a fraction into separate terms before integrating:
∫xx2+3x−1dx=∫(x+3−x1)dx=2x2+3x−log∣x∣+C.
Trigonometric expansion
Many trigonometric integrands have no direct formula in the form given, but do have one after a standard identity is used to expand them into a sum:
sin2x=21−cos2x,cos2x=21+cos2x
So, for example,
∫sin2xdx=∫21−cos2xdx=2x−4sin2x+C.
Product-to-sum identities do the same job for products such as sin3xcos5x, and sin3x, cos3x can be expanded using their triple-angle forms.
How to use it
- Look at the integrand — is it a product, a power, or a single fraction over x?
- Expand it (multiply out, or apply a trig identity) into a sum of standard terms. …
Writing the numerator one as sine-squared plus cosine-squared splits the single fraction into the sum of a secant-squared and a cosecant-squared term, each of which integrates to a standard result. …
Write 1=sin2x+cos2x in the numerator to split the integrand into two standard integrable pieces.
sin2xcos2x1=sin2xcos2xsin2x+cos2x=sin2xcos2xsin2x+sin2xcos2xcos2x=cos2x1+sin2x1=sec2x+csc2x.
So …
- CBSE 2026Set ANNUAL1 markQ.Evaluate \int \dfrac{2 - 3\sin x}{\cos^2 x},dx.
›Reveal solutionSolution
Split the integrand into two standard integrals, sec2x and secxtanx.
Working: Split the fraction:
∫cos2x2−3sinxdx=∫cos2x2dx−∫cos2x3sinxdx=2∫sec2xdx−3∫secxtanxdx
…
- CBSE 2026Set ANNUAL1 markQ.Find the integral: \int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 dx
›Reveal solutionSolution
∫(x−x1)2dx=2x2−2x+ln∣x∣+c.
Concept. Expand the integrand into simple power terms, then integrate term by term using ∫xndx=n+1xn+1 and ∫x1dx=ln∣x∣.
Steps. …
- CBSE 2024Set ANNUAL1 markQ.Evaluate ∫ cosec x (cosec x + cot x) dx.
›Reveal solutionSolution
Split the product and integrate each standard term separately.
∫cscx(cscx+cotx)dx=∫csc2xdx+∫cscxcotxdx
…
- CBSE 2023Set E1 markMCQQ.∫x(4x2−6)dx=(a) 4x3−6x+k(b) 34x4−6x2+k(c) x4−3x2+k(d) 34x3−3x2+k
›Reveal solutionSolution
∫x(4x2−6)dx=x4−3x2+k.
Expand the integrand: x(4x2−6)=4x3−6x. Integrate term by term:
…
- CBSE 2023Set ANNUAL1 markQ.Evaluate : ∫x2(1−x21)dx OR Evaluate : ∫tan2xdx
›Reveal solutionSolution
Simplify the integrand to a polynomial, then integrate term by term.
∫x2(1−x21)dx=∫(x2−1)dx=3x3−x+c.
…
- CBSE 2022Set ANNUAL1 markQ.Evaluate : ∫secx(secx+tanx)dx OR Evaluate : ∫23x1dx
›Reveal solutionSolution
Expand the integrand into standard forms whose antiderivatives are known.
∫secx(secx+tanx)dx=∫(sec2x+secxtanx)dx.
Using ∫sec2xdx=tanx and ∫secxtanxdx=secx,
=tanx+secx+C.
…
- CBSE 2019Set ANNUAL1 markQ.Find ∫(1 - x)√x · dx.
›Reveal solutionSolution
Write (1−x)x=x1/2−x3/2 and integrate: 32x3/2−52x5/2+C.
Concept. Power rule: ∫xndx=n+1xn+1+C (for ne−1).
Working.
(1−x)x=x−xx=x1/2−x3/2. …
- CBSE 2018Set ANNUAL1 markQ.Find ∫ (x^3 − 1)/x^2 dx.
›Reveal solutionSolution
∫x2x3−1dx=2x2+x1+C.
Concept. Simplify a rational integrand by term-by-term division, then use the power rule ∫xndx=n+1xn+1+C.
Step-by-step.
x2x3−1=x2x3−x21=x−x−2. …
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