Integrate: sin^2(2x+5).
Hint. Use the half-angle identity sin^2(theta) = (1-cos2theta)/2 with theta=2x+5.
Since sin^2(2x+5) = [1-cos(4x+10)]/2, integrating term by term gives x/2 for the constant piece and -(1/8)sin(4x+10) for the cosine piece, because the inner argument's derivative is 4.
✦ x/2 - (1/8) sin(4x+10) + C
Integrate: sin 3x cos 4x.
Hint. Use the product-to-sum identity sinA cosB = (1/2)[sin(A+B)+sin(A-B)].
Since sin3x cos4x = (1/2)[sin7x + sin(-x)] = (1/2)[sin7x - sinx], integrating each sine term separately gives the result.
✦ -(1/14) cos 7x + (1/2) cos x + C
Integrate: cos 2x cos 4x cos 6x.
Hint. Combine cos2x and cos6x into a sum first using the product-to-sum identity, then multiply by cos4x and combine again.
First, cos2x cos6x = (1/2)(cos8x+cos4x). Multiplying by cos4x and converting cos4x.cos8x and cos^2(4x) via product-to-sum identities gives a sum of cos12x, cos4x, cos8x, and a constant, each of which integrates to a sine term or a linear term.
✦ sin 12x/48 + sin 4x/16 + sin 8x/32 + x/4 + C
Integrate: sin^3(2x+1).
Hint. Use the identity sin^3(theta) = [3sin(theta) - sin(3theta)]/4 with theta=2x+1.
Since sin^3(2x+1) = [3sin(2x+1)-sin(6x+3)]/4, integrating each sine term (accounting for the chain-rule factor on each inner argument) gives the result.
✦ -(3/8) cos(2x+1) + (1/24) cos(6x+3) + C
Integrate: sin^3 x cos^3 x.
Hint. Write cos^3x = cos^2x . cosx = (1-sin^2x)cosx, then substitute t=sinx.
Writing sin^3x cos^3x = sin^3x(1-sin^2x)cosx, put t=sinx so dt=cosx dx, giving integral of t^3(1-t^2)dt = integral of (t^3-t^5)dt.
✦ sin^4 x/4 - sin^6 x/6 + C
Integrate: sin x sin 2x sin 3x.
Hint. Combine sinx and sin3x into a sum first using product-to-sum, then multiply by sin2x and combine again.
First, sinx sin3x = (1/2)(cos2x-cos4x). Multiplying by sin2x and converting each product to a sum via product-to-sum identities gives a combination of sin4x, sin6x, and sin2x terms, each integrating directly.
✦ -cos 4x/16 + cos 6x/24 - cos 2x/8 + C
Integrate: sin 4x sin 8x.
Hint. Use the product-to-sum identity sinA sinB = (1/2)[cos(A-B)-cos(A+B)].
Since sin4x sin8x = (1/2)[cos4x-cos12x], integrating each cosine term separately gives the result.
✦ sin 4x/8 - sin 12x/24 + C
Integrate: (1-cos x)/(1+cos x).
Hint. Use 1-cosx=2sin^2(x/2) and 1+cosx=2cos^2(x/2), so the ratio is tan^2(x/2).
The ratio simplifies to tan^2(x/2) = sec^2(x/2)-1, so integrating gives 2tan(x/2) from the secant-squared piece (the factor of 2 from the chain rule on x/2), minus x from the constant piece.
✦ 2 tan(x/2) - x + C
Integrate: cos x/(1+cos x).
Hint. Add and subtract 1 in the numerator to split off a constant piece: cosx = (1+cosx) - 1.
Writing cosx/(1+cosx) = 1 - 1/(1+cosx), and using 1+cosx=2cos^2(x/2) so 1/(1+cosx)=(1/2)sec^2(x/2), the integral becomes x minus tan(x/2).
✦ x - tan(x/2) + C
Integrate: sin^4 x.
Hint. Square the half-angle identity sin^2x=(1-cos2x)/2, then apply the half-angle identity again to the resulting cos^2(2x) term.
Expanding [(1-cos2x)/2]^2 = (1-2cos2x+cos^2(2x))/4, and rewriting cos^2(2x) via the half-angle identity again, gives a combination of a constant, cos2x, and cos4x terms, each of which integrates directly.
✦ 3x/8 - (1/4) sin 2x + (1/32) sin 4x + C
Integrate: cos^4 2x.
Hint. Use the identity cos^4(theta) = (3+4cos2theta+cos4theta)/8 with theta=2x.
Substituting theta=2x into cos^4(theta) = (3+4cos2theta+cos4theta)/8 gives (3+4cos4x+cos8x)/8, and integrating each of the three terms gives the result.
✦ 3x/8 + sin 4x/8 + sin 8x/64 + C
Integrate: sin^2 x/(1+cos x).
Hint. Factor the numerator using sin^2x = (1-cosx)(1+cosx), which cancels the denominator.
Since sin^2x = 1-cos^2x = (1-cosx)(1+cosx), the fraction simplifies to (1-cosx), which integrates term by term.
✦ x - sin x + C
Integrate: (cos 2x - cos 2 alpha)/(cos x - cos alpha).
Hint. Use cos2theta = 2cos^2(theta)-1 to rewrite the numerator as a difference of squares in cosx and cos(alpha).
Since cos2x-cos2alpha = 2cos^2x-2cos^2(alpha) = 2(cosx-cos alpha)(cosx+cos alpha), the (cosx-cos alpha) factor cancels with the denominator, leaving 2(cosx+cos alpha) to integrate.
✦ 2 sin x + 2x cos(alpha) + C
Integrate: (cos x - sin x)/(1+sin 2x).
Hint. Recognise 1+sin2x as a perfect square: (sinx+cosx)^2.
Since 1+sin2x = sin^2x+cos^2x+2sinxcosx = (sinx+cosx)^2, put t=sinx+cosx so dt=(cosx-sinx)dx, exactly the numerator, giving integral of dt/t^2 = -1/t.
✦ -1/(sin x + cos x) + C
Integrate: tan^3(2x) sec(2x).
Hint. Write tan^2(2x)=sec^2(2x)-1 and put t=sec2x, since d(sec2x)=2sec2x tan2x dx matches the remaining factors.
Writing tan^3(2x)sec(2x) = [sec^2(2x)-1] . tan(2x)sec(2x), put t=sec2x so dt=2sec2x tan2x dx, giving integral of (t^2-1)dt/2 = (1/2)(t^3/3-t).
✦ sec^3(2x)/6 - sec(2x)/2 + C
Integrate: tan^4 x.
Hint. Write tan^4x = tan^2x(sec^2x-1) and split into two integrals, using t=tanx for the first.
Since tan^4x = tan^2x sec^2x - tan^2x, the first piece integrates via t=tanx to tan^3x/3, and the second piece tan^2x=sec^2x-1 integrates to tanx-x; adding these gives the combined result.
✦ tan^3 x/3 - tan x + x + C
Integrate: (sin^3 x + cos^3 x)/(sin^2 x cos^2 x).
Hint. Split the fraction into two separate terms and simplify each using sin^2x+cos^2x cancellation.
Splitting, sin^3x/(sin^2xcos^2x) = sinx/cos^2x = secx tanx, and cos^3x/(sin^2xcos^2x) = cosx/sin^2x = cosecx cotx. Integrating secx tanx gives secx, and integrating cosecx cotx gives -cosecx.
✦ sec x - cosec x + C
Integrate: (cos 2x + 2sin^2 x)/cos^2 x.
Hint. Use cos2x=1-2sin^2x so the numerator collapses to a constant.
Since cos2x = 1-2sin^2x, the numerator cos2x+2sin^2x equals exactly 1, so the whole integrand reduces to 1/cos^2x = sec^2x, which integrates to tanx.
✦ tan x + C
Integrate: 1/(sin x cos^3 x).
Hint. Divide numerator and denominator by cos^4 x, turning this into a function of tan x alone.
Dividing through by cos^4x, the integrand becomes sec^4x/tanx = sec^2x(1+tan^2x)/tanx. With t=tanx, dt=sec^2xdx, this becomes (1/t+t)dt, which integrates to log|t|+t^2/2.
✦ log|tan x| + tan^2 x/2 + C
Integrate: cos 2x/(cos x + sin x)^2.
Hint. Factor cos2x = (cosx-sinx)(cosx+sinx) so one factor cancels with the square in the denominator.
Since cos2x = (cosx-sinx)(cosx+sinx), the fraction reduces to (cosx-sinx)/(cosx+sinx). Putting t=sinx+cosx, dt=(cosx-sinx)dx exactly matches the numerator, giving integral of dt/t = log|t|.
✦ log|sin x + cos x| + C
Integrate: sin^-1(cos x).
Hint. Use cosx = sin(pi/2 - x) to rewrite the inverse sine as a linear expression in x.
Since cosx = sin(pi/2-x), we have sin^-1(cosx) = pi/2-x on the relevant domain, so the integral reduces to integrating a linear function of x.
✦ (pi x)/2 - x^2/2 + C
Integrate: 1/(cos(x-a) cos(x-b)).
Hint. Multiply and divide by sin(a-b), then write 1 as sin[(x-b)-(x-a)]/sin(a-b) and expand using the sine subtraction formula.
Writing 1 = sin[(x-b)-(x-a)]/sin(a-b) and expanding via the sine-difference formula, the integrand splits into [tan(x-b)-tan(x-a)]/sin(a-b); since each tangent integrates to a log of a secant, this gives a difference of two logs.
✦ (1/sin(a-b)) log|cos(x-a)/cos(x-b)| + C
Choose the correct answer: integral of (sin^2 x - cos^2 x)/(sin^2 x cos^2 x) dx is equal to (A) tan x+cot x+C (B) tan x+cosec x+C (C) -tan x+cot x+C (D) tan x+sec x+C
Hint. Split the fraction into 1/cos^2x minus 1/sin^2x, i.e. sec^2x-cosec^2x, then integrate each piece.
Since (sin^2x-cos^2x)/(sin^2xcos^2x) = 1/cos^2x - 1/sin^2x = sec^2x-cosec^2x, integrating sec^2x gives tanx and integrating -cosec^2x gives cotx (since integral of cosec^2x is -cotx), so the sum is tanx+cotx.
✦ (A) tan x + cot x + C
Choose the correct answer: integral of e^x(1+x)/cos^2(xe^x) dx equals (A) -cot(xe^x)+C (B) tan(xe^x)+C (C) tan(e^x)+C (D) cot(e^x)+C
Hint. Put t=x e^x, and check that dt is exactly the numerator e^x(1+x)dx.
With t=x e^x, dt = (e^x + x e^x)dx = e^x(1+x)dx, exactly the numerator. Since cos^2(xe^x)=cos^2t, the integral becomes integral of sec^2t dt = tan t.
✦ (B) tan(x e^x) + C
Solutions written by the tuition.in editorial team and checked against lemh201.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.
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