Evaluate using substitution: integral from 0 to 1 of x/(x^2+1) dx.
Hint. Put t=x^2+1, so dt=2x dx, and change the limits to match t instead of x.
With t=x^2+1, dt=2xdx; when x=0, t=1, and when x=1, t=2. The integral becomes (1/2) integral from 1 to 2 of dt/t = (1/2)[log t] evaluated at the new limits.
✦ (1/2) log 2
Evaluate using substitution: integral from 0 to pi/2 of sqrt(sin phi) cos^5(phi) dphi.
Hint. Write cos^5(phi) as (1-sin^2phi)^2 cosphi, then put t=sin phi to change the limits.
With t=sinphi, dt=cosphi dphi; since phi=0 gives t=0 and phi=pi/2 gives t=1, the integral becomes integral from 0 to 1 of sqrt(t)(1-t^2)^2 dt, a polynomial in t^(1/2) that integrates term by term.
✦ 64/231
Evaluate using substitution: integral from 0 to 1 of sin^-1(2x/(1+x^2)) dx.
Hint. Recognise 2x/(1+x^2) as sin(2 tan^-1 x), simplifying the inverse sine to a linear multiple of tan^-1x.
Since sin^-1(2x/(1+x^2))=2tan^-1x for |x| less than or equal to 1, the integral becomes 2 integral from 0 to 1 of tan^-1x dx, which integrates by parts to 2[x tan^-1x - (1/2)log(1+x^2)].
✦ pi/2 - log 2
Evaluate using substitution: integral from 0 to 2 of x sqrt(x+2) dx. [Put x+2=t^2]
Hint. Following the hint, put x+2=t^2 so x=t^2-2 and dx=2t dt, and change the limits to match t.
With t=sqrt(x+2), x=t^2-2, dx=2tdt; when x=0, t=sqrt(2), and when x=2, t=2. The integral becomes 2 integral from sqrt(2) to 2 of (t^2-2)t^2 dt, a polynomial in t.
✦ (32 + 16 sqrt(2))/15
Evaluate using substitution: integral from 0 to pi/2 of sin x/(1+cos^2 x) dx.
Hint. Put t=cos x, so dt=-sin x dx, and change the limits to match t.
With t=cosx, dt=-sinxdx; when x=0, t=1, and when x=pi/2, t=0. The integral becomes integral from 0 to 1 of dt/(1+t^2) (the sign flip and limit swap cancel), which is a standard tan^-1 form.
✦ pi/4
Evaluate using substitution: integral from 0 to 2 of dx/(x+4-x^2).
Hint. Complete the square in the denominator, then apply the standard log-ratio formula for a quadratic denominator.
Completing the square, x+4-x^2 = 17/4-(x-1/2)^2, so this matches the standard form 1/(a^2-u^2), giving a log-ratio antiderivative; evaluating at both limits gives the numeric result.
✦ (1/sqrt(17)) log[(3+sqrt(17))/(sqrt(17)-3)] - (1/sqrt(17)) log[(sqrt(17)-1)/(sqrt(17)+1)]
Evaluate using substitution: integral from -1 to 1 of dx/(x^2+2x+5).
Hint. Complete the square: x^2+2x+5=(x+1)^2+4, matching the standard tan^-1 form.
Completing the square gives (x+1)^2+4, so the antiderivative is (1/2)tan^-1((x+1)/2). Evaluating at x=1 (giving pi/8) and x=-1 (giving 0) and subtracting gives the result.
✦ pi/8
Evaluate using substitution: integral from 1 to 2 of (1/x - 1/(2x^2)) e^(2x) dx.
Hint. Try to find h(x) such that 2h(x)+h'(x) matches the integrand, since d/dx[e^(2x)h(x)] = e^(2x)[2h(x)+h'(x)].
Trying h(x)=1/(2x) gives h'(x)=-1/(2x^2), so 2h(x)+h'(x)=1/x-1/(2x^2), exactly the integrand. So the antiderivative is e^(2x)/(2x); evaluating at x=2 and x=1 and subtracting gives the result.
✦ e^4/4 - e^2/2
Choose the correct answer: the value of the integral from 1/3 to 1 of [(x-x^3)^(1/3)]/x^4 dx is (A) 6 (B) 0 (C) 3 (D) 4
Hint. Divide numerator and denominator inside the cube root by a suitable power of x to reduce this to a substitution in t=1/x^2 or similar.
Writing (x-x^3)^(1/3)/x^4 = [(1-x^2)/x^2]^(1/3)/x^2 and substituting t=1/x^2-1 (equivalently (1-x^2)/x^2) reduces the integral to a simple power of t, and evaluating at the transformed limits gives 6.
✦ (A) 6
Choose the correct answer: if f(x) = integral from 0 to x of t sin t dt, then f'(x) is (A) cos x+x sin x (B) x sin x (C) x cos x (D) sin x+x cos x
Hint. Apply the First Fundamental Theorem of Calculus directly — no integration is needed.
By the First Fundamental Theorem of Calculus, f'(x) equals the integrand evaluated at x, i.e. f'(x) = x sin x, with no further work required.
✦ (B) x sin x
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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