Integrate: 1/(x-x^3).
Hint. Factor the denominator completely as x(1-x)(1+x), then decompose into three partial fractions.
Decomposing 1/[x(1-x)(1+x)] into A/x+B/(1-x)+C/(1+x) and substituting the roots gives A=1, B=1/2, C=-1/2, so the integral combines into a log-ratio form.
✦ log|x| - (1/2) log|1-x^2| + C
Integrate: 1/(sqrt(x+a)+sqrt(x+b)).
Hint. Rationalise by multiplying numerator and denominator by (sqrt(x+a)-sqrt(x+b)) — the denominator becomes the constant a-b.
Since multiplying by the conjugate turns the denominator (x+a)-(x+b) into the constant a-b, this leaves integral of [sqrt(x+a)-sqrt(x+b)]/(a-b) dx, which integrates term by term via the power rule.
✦ (2/(3(a-b))) [(x+a)^(3/2) - (x+b)^(3/2)] + C
Integrate: 1/(x sqrt(ax-x^2)). [Hint: put x=a/t]
Hint. Following the hint, put x=a/t so dx=-a/t^2 dt, turning the square root into a simpler expression in t.
With x=a/t, ax-x^2 simplifies to a^2(t-1)/t^2, so sqrt(ax-x^2)=a sqrt(t-1)/t, and after substitution the integral reduces to a simple power of (t-1), giving -(2/a)sqrt(t-1), which converts back to x.
✦ -(2/a) sqrt((a-x)/x) + C
Integrate: 1/[x^2 (x^4+1)^(3/4)].
Hint. Factor x^4 out of (x^4+1) so the expression becomes a function of (1+1/x^4), then substitute t=1+1/x^4.
Writing (x^4+1)^(3/4)=x^3(1+x^-4)^(3/4), the integrand becomes x^-5(1+x^-4)^(-3/4). With t=1+x^-4, dt=-4x^-5dx, the integral becomes a simple power of t.
✦ -(x^4+1)^(1/4)/x + C
Integrate: 1/(x^(1/2)+x^(1/3)). [Hint: rewrite as 1/[x^(1/3)(1+x^(1/6))], put x=t^6]
Hint. Following the hint, put x=t^6 so both the square root and cube root become whole powers of t, turning the denominator into t^2(t+1).
With x=t^6, dx=6t^5dt, the denominator x^(1/2)+x^(1/3)=t^3+t^2=t^2(t+1), so the integral becomes 6 integral of t^3/(t+1)dt, and dividing t^3 by (t+1) leaves a polynomial plus a log term.
✦ 2 sqrt(x) - 3x^(1/3) + 6x^(1/6) - 6 log(x^(1/6)+1) + C
Integrate: 5x/[(x+1)(x^2+9)].
Hint. Decompose into A/(x+1) + (Bx+C)/(x^2+9), finding A by substituting x=-1 and B, C by matching coefficients.
Substituting x=-1 gives A=-1/2, and matching the x^2 and constant coefficients gives B=1/2, C=9/2. Integrating the linear-over-quadratic piece splits into a log and a tan^-1 term.
✦ -(1/2) log|x+1| + (1/4) log(x^2+9) + (3/2) tan^-1(x/3) + C
Integrate: sin x/sin(x-a).
Hint. Expand sinx as sin[(x-a)+a] using the sine addition formula, then split into a constant piece and a cot(x-a) piece.
Writing sinx=sin(x-a)cosa+cos(x-a)sina, dividing through by sin(x-a) gives cosa + sina.cot(x-a), and integrating gives a linear term from the constant plus a log term from the cotangent.
✦ x cos(a) + sin(a) log|sin(x-a)| + C
Integrate: (e^(5 log x) - e^(4 log x))/(e^(3 log x) - e^(2 log x)).
Hint. Simplify each e^(k log x) term to x^k first — the expression collapses to a simple polynomial.
Since e^(k logx)=x^k, the expression simplifies to (x^5-x^4)/(x^3-x^2) = x^4(x-1)/[x^2(x-1)] = x^2 after cancelling, which integrates directly by the power rule.
✦ x^3/3 + C
Integrate: cos x/sqrt(4-sin^2 x).
Hint. Put t=sin x, so dt=cos x dx, reducing this to the standard sin^-1 form.
With t=sinx, dt=cosxdx, the integral becomes integral of dt/sqrt(4-t^2), the standard form sin^-1(t/2).
✦ sin^-1(sin x/2) + C
Integrate: (sin^8 x - cos^8 x)/(1-2 sin^2 x cos^2 x).
Hint. Factor the numerator as a difference of squares twice, and notice it shares a common factor with the denominator that cancels completely.
Factoring, sin^8x-cos^8x = (sin^2x-cos^2x)(sin^4x+cos^4x) = -cos2x.(1-2sin^2xcos^2x), so the (1-2sin^2xcos^2x) factor cancels exactly with the denominator, leaving just -cos2x to integrate.
✦ -sin(2x)/2 + C
Integrate: 1/[cos(x+a) cos(x+b)].
Hint. Multiply and divide by sin(b-a), then write 1 as sin[(x+b)-(x+a)]/sin(b-a) and expand using the sine subtraction formula.
Writing 1 = sin[(x+b)-(x+a)]/sin(b-a) and expanding, the integrand splits into [tan(x+b)-tan(x+a)]/sin(b-a); since each tangent integrates to a log of a cosine, this gives a difference of two logs.
✦ (1/sin(a-b)) log|cos(x+b)/cos(x+a)| + C
Integrate: x^3/sqrt(1-x^8).
Hint. Put t=x^4, so dt=4x^3 dx, reducing this to the standard sin^-1 form.
With t=x^4, dt=4x^3dx, the integral becomes (1/4) integral of dt/sqrt(1-t^2), the standard form sin^-1(t).
✦ (1/4) sin^-1(x^4) + C
Integrate: e^x/[(1+e^x)(2+e^x)].
Hint. Put t=e^x, so dt=e^x dx, reducing this to a rational function of t that decomposes into partial fractions.
With t=e^x, dt=e^xdx, the integral becomes integral of dt/[(1+t)(2+t)], which decomposes into 1/(1+t) - 1/(2+t), each integrating to a log.
✦ log[(1+e^x)/(2+e^x)] + C
Integrate: 1/[(x^2+1)(x^2+4)].
Hint. Decompose into A/(x^2+1)+B/(x^2+4) treating x^2 as the variable, then match coefficients.
Writing 1=A(x^2+4)+B(x^2+1) and matching coefficients gives A=1/3, B=-1/3, so the integral is a combination of two tan^-1 terms.
✦ (1/3) tan^-1 x - (1/6) tan^-1(x/2) + C
Integrate: cos^3 x . e^(log sin x).
Hint. Simplify e^(log sin x) to sin x first, then use the substitution t=cos x on the resulting product.
Since e^(log sinx)=sinx, the integrand becomes cos^3x sinx. With t=cosx, dt=-sinxdx, the integral becomes -integral of t^3 dt.
✦ -cos^4 x/4 + C
Integrate: e^(3 log x) (x^4+1)^-1.
Hint. Simplify e^(3 log x) to x^3 first, then substitute t=x^4+1.
Since e^(3logx)=x^3, the integrand becomes x^3/(x^4+1). With t=x^4+1, dt=4x^3dx, the integral becomes (1/4) integral of dt/t.
✦ (1/4) log(x^4+1) + C
Integrate: f'(ax+b) [f(ax+b)]^n.
Hint. Put t=f(ax+b), so dt=a.f'(ax+b) dx by the chain rule, reducing this to a pure power of t.
With t=f(ax+b), dt=a f'(ax+b)dx, the integral becomes (1/a) integral of t^n dt = (1/a).t^(n+1)/(n+1), giving the answer once t is replaced back.
✦ [f(ax+b)]^(n+1) / [a(n+1)] + C
Integrate: 1/sqrt(sin^3 x sin(x+alpha)).
Hint. Divide inside the square root by sin^4x to rewrite everything in terms of cot x, then substitute for the resulting expression.
Dividing sin^3x sin(x+alpha) by sin^4x gives sin(x+alpha)/sinx = cos(alpha)+sin(alpha)cot(x), so the integrand becomes cosec^2x/sqrt(cos(alpha)+sin(alpha)cot x); putting t equal to that expression under the root reduces it to a power of t.
✦ -(2/sin(alpha)) sqrt(cos(alpha) + sin(alpha) cot x) + C
Integrate: sqrt[(1-sqrt(x))/(1+sqrt(x))].
Hint. Multiply inside the root by (1-sqrt(x))/(1-sqrt(x)) to turn the denominator into 1-x, a perfect square times a factor.
Multiplying inside the root by (1-sqrt(x))/(1-sqrt(x)) turns the expression into (1-sqrt(x))^2/(1-x), so the square root simplifies to (1-sqrt(x))/sqrt(1-x). Splitting into two pieces and using x=sin^2(theta) on the second piece completes the integral.
✦ -2 sqrt(1-x) - sin^-1(sqrt(x)) + sqrt(x-x^2) + C
Integrate: (2+sin 2x)/(1+cos 2x) . e^x.
Hint. Simplify the trigonometric fraction using 1+cos2x=2cos^2x and sin2x=2sinxcosx first, then recognise the e^x[f(x)+f'(x)] pattern.
Simplifying, (2+sin2x)/(1+cos2x) = sec^2x+tanx, which matches f(x)+f'(x) exactly for f(x)=tanx, so the antiderivative is e^x f(x) = e^x tanx.
✦ e^x tan x + C
Integrate: (x^2+x+1)/[(x+1)^2 (x+2)].
Hint. Decompose into A/(x+1)+B/(x+1)^2+C/(x+2), since (x+1) is a repeated factor.
Clearing denominators and substituting x=-1 gives B=1, x=-2 gives C=3, and matching the x^2 coefficient gives A=-2. The B/(x+1)^2 term integrates to -B/(x+1).
✦ -2 log|x+1| - 1/(x+1) + 3 log|x+2| + C
Integrate: tan^-1 sqrt[(1-x)/(1+x)].
Hint. Substitute x=cos(theta), which turns (1-x)/(1+x) into tan^2(theta/2) by the half-angle identities.
With x=cos(theta), (1-x)/(1+x)=tan^2(theta/2), so the inverse tangent simplifies to theta/2 = (1/2)cos^-1x. Integrating this scaled inverse cosine using the standard by-parts result for integral of cos^-1x completes the problem.
✦ (x/2) cos^-1 x - (1/2) sqrt(1-x^2) + C
Integrate: sqrt(x^2+1)[log(x^2+1)-2log x]/x^4.
Hint. Simplify the log bracket to log(1+1/x^2) first, then substitute t=sqrt(x^2+1)/x, which turns both the square root and the log into powers and a log of t.
Since log(x^2+1)-2logx = log(1+1/x^2) = log(t^2) = 2log(t) with t=sqrt(x^2+1)/x, and dt=-dx/(x^3t), the integral becomes -2 integral of t^2 log(t) dt, a standard by-parts integral in t.
✦ -(2/3) t^3 log t + (2/9) t^3 + C, where t = sqrt(x^2+1)/x
Evaluate: integral from pi/2 to pi of e^x (1-sin x)/(1-cos x) dx.
Hint. Simplify the trigonometric fraction using half-angle identities to recognise the e^x[f(x)+f'(x)] pattern with f(x)=-cot(x/2).
Since 1-cosx=2sin^2(x/2) and sinx=2sin(x/2)cos(x/2), the fraction simplifies to (1/2)cosec^2(x/2)-cot(x/2), matching f(x)+f'(x) for f(x)=-cot(x/2), so the antiderivative e^x.(-cot(x/2)) evaluates to 0 at x=pi (cot(pi/2)=0) and -e^(pi/2) at x=pi/2.
✦ e^(pi/2)
Evaluate: integral from 0 to pi/4 of sin x cos x/(cos^4 x+sin^4 x) dx.
Hint. Put t=sin^2 x, so dt=2 sin x cos x dx, and rewrite the denominator in terms of t.
Since t=sin^2x gives dt=2sinxcosxdx and cos^4x+sin^4x=1-2t(1-t)=2t^2-2t+1, the integral becomes (1/2) integral of dt/(2t^2-2t+1) from t=0 to t=1/2, a standard tan^-1 form after completing the square.
✦ pi/8
Evaluate: integral from 0 to pi/2 of cos^2 x/(cos^2 x+4 sin^2 x) dx.
Hint. Divide numerator and denominator by cos^2 x to rewrite everything in terms of tan x, then substitute t=tan x.
Since dividing by cos^2x turns the integrand into 1/(1+4tan^2x), putting t=tanx (so dt=(1+t^2)dx) gives a rational function of t that decomposes into partial fractions over (1+t^2) and (1+4t^2).
✦ pi/6
Evaluate: integral from pi/6 to pi/3 of (sin x+cos x)/sqrt(sin 2x) dx.
Hint. Put t=sin x - cos x, and note that t^2=1-sin2x, which directly matches the expression under the square root.
With t=sinx-cosx, dt=(cosx+sinx)dx exactly matches the numerator, and t^2=1-sin2x, so sin2x=1-t^2, turning the integral into the standard form integral of dt/sqrt(1-t^2) = sin^-1(t).
✦ 2 sin^-1((sqrt(3)-1)/2)
Evaluate: integral from 0 to 1 of dx/(sqrt(1+x)-sqrt(x)).
Hint. Rationalise by multiplying numerator and denominator by (sqrt(1+x)+sqrt(x)) — the denominator becomes exactly 1.
Because multiplying by the conjugate turns the denominator (1+x)-x into exactly 1, this leaves integral of [sqrt(1+x)+sqrt(x)]dx, which integrates term by term via the power rule and is then evaluated at the two limits.
✦ 4 sqrt(2)/3
Evaluate: integral from 0 to pi/4 of (sin x+cos x)/(9+16 sin 2x) dx.
Hint. Put t=sin x - cos x, and use 9+16sin2x = 25-16t^2, matching a standard log-ratio form.
Putting t=sinx-cosx gives dt=(cosx+sinx)dx, exactly the numerator, and since sin2x=1-t^2, the denominator becomes 25-16t^2, matching the standard log-ratio antiderivative; evaluating at the transformed limits t=-1 and t=0 gives the result.
✦ (1/20) log 3
Evaluate: integral from 0 to pi/2 of sin 2x tan^-1(sin x) dx.
Hint. Use integration by parts with u=tan^-1(sin x) and dv=sin2x dx=2 sinx cosx dx, noting v=sin^2x.
With u=tan^-1(sinx), dv=sin2xdx, v=sin^2x, integration by parts gives [sin^2x tan^-1(sinx)] evaluated at the limits (giving pi/4) minus integral of sin^2x cosx/(1+sin^2x)dx, which reduces via t=sinx to a standard rational integral evaluating to 1-pi/4.
✦ pi/2 - 1
Evaluate: integral from 1 to 4 of [|x-1|+|x-2|+|x-3|] dx.
Hint. Split the interval at the three points x=1,2,3 where the absolute values change sign, and integrate each of the three resulting linear pieces separately.
On [1,2], the sum simplifies to 4-x; on [2,3], to x; on [3,4], to 3x-6. Integrating each linear piece over its sub-interval and adding the three results gives the total.
✦ 9.5
Prove that: integral from 1 to 3 of dx/[x^2(x+1)] = 2/3 + log(2/3).
Hint. Decompose 1/[x^2(x+1)] into partial fractions first, then integrate and evaluate at the two limits.
Decomposing into 1/x^2 - 1/x + 1/(x+1), the antiderivative is -1/x - log|x| + log|x+1| = -1/x + log|(x+1)/x|. Evaluating at x=3 and x=1 and subtracting gives exactly 2/3+log(2/3).
✦ Proved: 2/3 + log(2/3)
Prove that: integral from 0 to 1 of x e^x dx = 1.
Hint. Integrate by parts with u=x, dv=e^x dx, then evaluate at the two limits.
With u=x, dv=e^xdx, v=e^x, integration by parts gives [x e^x - e^x] evaluated from 0 to 1, which is (e-e)-(0-1)=0-(-1)=1, exactly as required.
✦ Proved: 1
Prove that: integral from -1 to 1 of x^17 cos^4 x dx = 0.
Hint. Recognise that x^17 cos^4x is an odd function (odd power of x times an even function), so its integral over a symmetric interval vanishes.
Since x^17 is odd and cos^4x is even, their product is odd, i.e. f(-x)=-f(x). By property P7, the integral of an odd function over a symmetric interval [-a,a] is exactly zero, with no computation needed.
✦ Proved: 0
Prove that: integral from 0 to pi/2 of sin^3 x dx = 2/3.
Hint. Write sin^3x = sin^2x . sinx = (1-cos^2x)sinx, then substitute t=cos x.
With t=cosx, dt=-sinxdx, the integral becomes integral of (1-t^2)dt from t=1 to t=0 (with the sign flip from the limit swap), evaluating to 2/3 exactly.
✦ Proved: 2/3
Prove that: integral from 0 to pi/4 of 2 tan^3 x dx = 1 - log 2.
Hint. Write tan^3x = tan x . tan^2x = tan x(sec^2x-1), splitting into two separate integrals.
Splitting, 2tan^3x = 2tanx sec^2x - 2tanx. The first piece integrates (via t=tanx) to tan^2x, and the second integrates to 2log|cosx|. Evaluating both at pi/4 and 0 and combining gives exactly 1-log2.
✦ Proved: 1 - log 2
Prove that: integral from 0 to 1 of sin^-1 x dx = pi/2 - 1.
Hint. Integrate by parts with u=sin^-1x, dv=dx, then evaluate the resulting integral of x/sqrt(1-x^2) via a simple substitution.
With u=sin^-1x, dv=dx, v=x, integration by parts gives [x sin^-1x] evaluated at the limits (giving pi/2) minus integral of x/sqrt(1-x^2)dx, which evaluates to 1 via t=1-x^2; the difference is exactly pi/2-1.
✦ Proved: pi/2 - 1
Choose the correct answer: integral of dx/(e^x+e^-x) is equal to (A) tan^-1(e^x)+C (B) tan^-1(e^-x)+C (C) log(e^x-e^-x)+C (D) log(e^x+e^-x)+C
Hint. Multiply numerator and denominator by e^x, then substitute t=e^x to reduce this to a standard tan^-1 form.
Multiplying by e^x/e^x, the integrand becomes e^x/(e^(2x)+1). With t=e^x, dt=e^xdx, the integral becomes integral of dt/(1+t^2) = tan^-1(t) = tan^-1(e^x).
✦ (A) tan^-1(e^x) + C
Choose the correct answer: integral of cos 2x/(sin x+cos x)^2 dx is equal to (A) -1/(sin x+cos x)+C (B) log|sin x+cos x|+C (C) log|sin x-cos x|+C (D) 1/(sin x+cos x)^2
Hint. Factor cos2x = (cos x-sin x)(cos x+sin x), so one factor cancels with the square in the denominator.
Since cos2x=(cosx-sinx)(cosx+sinx), the fraction reduces to (cosx-sinx)/(sinx+cosx). Putting t=sinx+cosx, dt=(cosx-sinx)dx exactly matches the numerator, giving integral of dt/t = log|t|.
✦ (B) log|sin x + cos x| + C
Choose the correct answer: if f(a+b-x)=f(x), then integral from a to b of x f(x) dx is equal to (A) (a+b)/2 integral from a to b of f(b-x)dx (B) (a+b)/2 integral from a to b of f(b+x)dx (C) (b-a)/2 integral from a to b of f(x)dx (D) (a+b)/2 integral from a to b of f(x)dx
Hint. Apply property P4 (replace x with a+b-x), use the given symmetry f(a+b-x)=f(x), and add the transformed integral to the original.
Let I be the integral. Applying x to a+b-x and using f(a+b-x)=f(x) turns x f(x) into (a+b-x)f(x). Adding this to the original integrand xf(x) gives (a+b)f(x), so 2I = (a+b) integral of f(x)dx, giving the stated form.
✦ (D) (a+b)/2 integral from a to b of f(x) dx
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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