Using properties of definite integrals, evaluate: integral from 0 to pi/2 of cos^2 x dx.
Hint. Rewrite cos^2x=(1+cos2x)/2 and integrate directly — no property trick is needed here.
Using cos^2x=(1+cos2x)/2, the antiderivative is x/2+(1/4)sin2x, which evaluates to pi/4 at x=pi/2 (since sin(pi)=0) and 0 at x=0.
✦ pi/4
Using properties of definite integrals, evaluate: integral from 0 to pi/2 of sqrt(sin x)/(sqrt(sin x)+sqrt(cos x)) dx.
Hint. Apply property P4 (replace x with pi/2-x) — the transformed integrand is the complementary fraction, so adding it to the original gives 1.
Let I be the integral. Applying x to pi/2-x swaps sinx and cosx, turning the integrand into sqrt(cosx)/(sqrt(cosx)+sqrt(sinx)), which is exactly 1 minus the original integrand. So I = pi/2 - I, giving 2I=pi/2.
✦ pi/4
Using properties of definite integrals, evaluate: integral from 0 to pi/2 of sin^(3/2)x/(sin^(3/2)x+cos^(3/2)x) dx.
Hint. Apply property P4 (replace x with pi/2-x) — the same symmetry trick as the previous question works here too.
Call the integral I. Since applying x to pi/2-x swaps the roles of sin and cos, the integrand turns into cos^(3/2)x/(cos^(3/2)x+sin^(3/2)x), which is 1 minus the original integrand, so I=pi/2-I, giving 2I=pi/2.
✦ pi/4
Using properties of definite integrals, evaluate: integral from 0 to pi/2 of cos^5 x/(sin^5 x+cos^5 x) dx.
Hint. Apply property P4 (replace x with pi/2-x) — the same symmetry trick applies once more.
Denote the integral by I. Applying x to pi/2-x again swaps sin and cos, turning the integrand into sin^5x/(cos^5x+sin^5x), which is 1 minus the original integrand, so I=pi/2-I, giving 2I=pi/2.
✦ pi/4
Using properties of definite integrals, evaluate: integral from -5 to 5 of |x+2| dx.
Hint. Split the interval at x=-2, where the expression inside the absolute value changes sign, then integrate each piece separately.
For x less than -2, |x+2|=-(x+2), and for x greater than -2, |x+2|=x+2. Splitting the integral at x=-2 and evaluating each piece as a simple quadratic antiderivative gives the total.
✦ 29
Using properties of definite integrals, evaluate: integral from 2 to 8 of |x-5| dx.
Hint. Split the interval at x=5, where the expression inside the absolute value changes sign, then integrate each piece separately.
For x less than 5, |x-5|=-(x-5), and for x greater than 5, |x-5|=x-5. Splitting the integral at x=5 and evaluating each piece as a simple quadratic antiderivative gives the total.
✦ 9
Using properties of definite integrals, evaluate: integral from 0 to 1 of x(1-x)^n dx.
Hint. Apply property P4 (replace x with 1-x over the interval [0,1]) to turn x(1-x)^n into a simpler expression involving only powers of x.
Applying x to 1-x turns the integrand x(1-x)^n into (1-x)x^n = x^n-x^(n+1), which integrates term by term via the power rule to 1/(n+1)-1/(n+2).
✦ 1/[(n+1)(n+2)]
Using properties of definite integrals, evaluate: integral from 0 to pi/4 of log(1+tan x) dx.
Hint. Apply property P4 (replace x with pi/4-x) and use tan(pi/4-x) = (1-tanx)/(1+tanx) to simplify.
Let I be the integral. Applying x to pi/4-x and using the tangent subtraction formula, 1+tan(pi/4-x) simplifies to 2/(1+tanx), so log(1+tan(pi/4-x)) = log2 - log(1+tanx). This gives I = (pi/4)log2 - I, so 2I=(pi/4)log2.
✦ (pi/8) log 2
Using properties of definite integrals, evaluate: integral from 0 to 2 of x sqrt(2-x) dx.
Hint. Apply property P4 (replace x with 2-x) to turn this into an integral of (2-x)sqrt(x), then combine with the original.
Applying x to 2-x turns x sqrt(2-x) into (2-x)sqrt(x) = 2sqrt(x)-x^(3/2). Since this is a genuine transformation (not immediately giving I=const-I), integrating this form directly via the power rule gives the value.
✦ 16 sqrt(2)/15
Using properties of definite integrals, evaluate: integral from 0 to pi/2 of (2 log sin x - log sin 2x) dx.
Hint. Use sin2x=2sinxcosx to rewrite the integrand purely in terms of log(tan x) and a constant.
Since sin2x=2sinxcosx, we get 2logsinx-logsin2x = logsinx-logcosx-log2 = log(tanx)-log2. Since integral from 0 to pi/2 of log(tanx)dx=0 (by the x to pi/2-x symmetry, since tan(pi/2-x)=cotx=1/tanx flips the sign of the log), only the -log2 piece survives.
✦ -(pi/2) log 2
Using properties of definite integrals, evaluate: integral from -pi/2 to pi/2 of sin^2 x dx.
Hint. Recognise sin^2x is an even function, so the integral over the symmetric interval is twice the integral over half of it.
Since sin^2x is even (sin^2(-x)=sin^2x), property P7 gives integral from -pi/2 to pi/2 = 2 times integral from 0 to pi/2 of sin^2x dx = 2(pi/4).
✦ pi/2
Using properties of definite integrals, evaluate: integral from 0 to pi of x/(1+sin x) dx.
Hint. Apply property P4 (replace x with pi-x) to get an equation relating the integral to itself plus a simpler integral of 1/(1+sinx).
Let I be the integral. Applying x to pi-x (noting sin(pi-x)=sinx) gives I = integral of (pi-x)/(1+sinx)dx = pi.J - I, where J=integral from 0 to pi of dx/(1+sinx). Evaluating J=2 (via sec^2x-secxtanx split around the pi/2 discontinuity) gives 2I=2pi.
✦ pi
Using properties of definite integrals, evaluate: integral from -pi/2 to pi/2 of sin^7 x dx.
Hint. Recognise sin^7x is an odd function, so the integral over the symmetric interval vanishes.
Since sin^7x is odd (sin^7(-x)=-sin^7x), property P7 gives integral from -pi/2 to pi/2 of an odd function is exactly zero, with no computation needed.
✦ 0
Using properties of definite integrals, evaluate: integral from 0 to 2pi of cos^5 x dx.
Hint. Split the interval into [0,pi] and [pi,2pi], and use cos(x+pi)=-cosx to relate the second half to the first.
On [pi,2pi], substituting x=pi+u shows cos^5x=-cos^5u, so integral from pi to 2pi of cos^5x dx equals the negative of integral from 0 to pi of cos^5x dx. Adding the two halves cancels exactly to zero.
✦ 0
Using properties of definite integrals, evaluate: integral from 0 to pi/2 of (sin x - cos x)/(1+sin x cos x) dx.
Hint. Apply property P4 (replace x with pi/2-x) — the numerator flips sign while the denominator stays the same.
Writing I for the integral, since applying x to pi/2-x swaps sinx and cosx, the numerator sinx-cosx turns into cosx-sinx=-(sinx-cosx), while sinxcosx (and hence the denominator) is unchanged, so I=-I, giving 2I=0.
✦ 0
Using properties of definite integrals, evaluate: integral from 0 to pi of log(1+cos x) dx.
Hint. Use 1+cosx=2cos^2(x/2) to split the log into a constant plus a log-cosine term, then use the known result for integral of log(cos) over this kind of interval.
Writing log(1+cosx)=log2+2log|cos(x/2)|, the integral splits into pi.log2 plus 2 times integral from 0 to pi of log(cos(x/2))dx. Evaluating the cosine-log piece (a standard result related to the Fourier series of log(cosine)) gives a total of -pi.log2.
✦ -pi log 2
Using properties of definite integrals, evaluate: integral from 0 to a of sqrt(x)/(sqrt(x)+sqrt(a-x)) dx.
Hint. Apply property P4 (replace x with a-x) — adding the transformed integral to the original gives a constant integrand of 1.
Let I be the integral. Applying x to a-x turns the integrand into sqrt(a-x)/(sqrt(a-x)+sqrt(x)), and adding this to the original integrand gives exactly 1 for every x. So 2I = integral from 0 to a of 1 dx = a.
✦ a/2
Using properties of definite integrals, evaluate: integral from 0 to 4 of |x-1| dx.
Hint. Split the interval at x=1, where the expression inside the absolute value changes sign, then integrate each piece separately.
For x less than 1, |x-1|=-(x-1), and for x greater than 1, |x-1|=x-1. Splitting the integral at x=1 and evaluating each piece as a simple quadratic antiderivative gives the total.
✦ 5
Show that: integral from 0 to a of f(x)g(x) dx = 2 integral from 0 to a of f(x) dx, if f and g are defined as f(x)=f(a-x) and g(x)+g(a-x)=4.
Hint. Apply property P4 to the integral of f(x)g(x), then use both given conditions to combine the original and transformed integrals.
Let I = integral from 0 to a of f(x)g(x)dx. Applying x to a-x and using f(a-x)=f(x) gives I = integral from 0 to a of f(x)g(a-x)dx. Adding this to the original I gives 2I = integral from 0 to a of f(x)[g(x)+g(a-x)]dx = integral from 0 to a of f(x).4 dx, since g(x)+g(a-x)=4 for every x. So 2I = 4 integral from 0 to a of f(x)dx, giving I = 2 integral from 0 to a of f(x)dx as required.
✦ Proved: I = 2 integral from 0 to a of f(x) dx
Choose the correct answer: the value of the integral from -pi/2 to pi/2 of (x^3+x cos x+tan^5 x+1) dx is (A) 0 (B) 2 (C) pi (D) 1
Hint. Check which terms are odd functions (which vanish over the symmetric interval) and which term is even.
x^3, x cosx, and tan^5x are all odd functions (each picks up a minus sign under x to -x), so by property P7 each contributes zero. Only the constant term 1 survives, contributing integral from -pi/2 to pi/2 of 1 dx = pi.
✦ (C) pi
Choose the correct answer: the value of the integral from 0 to pi/2 of log[(4+3sin x)/(4+3cos x)] dx is (A) 2 (B) 3/4 (C) 0 (D) -2
Hint. Apply property P4 (replace x with pi/2-x) — the fraction inside the log flips upside down, turning the log into its own negative.
Setting I equal to the integral, since applying x to pi/2-x swaps sinx and cosx, the fraction (4+3sinx)/(4+3cosx) turns into its reciprocal, so the log picks up a minus sign and I becomes -I, giving 2I=0.
✦ (C) 0
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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