Previous year questions: Integrals

Integrals32 min readLesson 8 of 8

Q1 (CBSE Board 2026, 3 marks) Evaluate the integral from 0 to 1 of x times the inverse tangent of x, with respect to x. Answer: Before integrating, note that the product x times inverse tangent is a classic by-parts pair: the inverse tangent is the logarithmic-type factor in the LIATE list, so it is chosen as u, and the algebraic x is integrated. Use by parts with u equal to the inverse tangent of x and v prime equal to x. Then v is half x squared, and the integral equals half x squared inverse tangent x evaluated from 0 to 1 minus half the integral from 0 to 1 of x squared over one plus x squared. The boundary term at 1 gives pi by eight and at 0 gives zero. Writing x squared over one plus x squared as one minus one over one plus x squared, the remaining integral is x minus inverse tangent x from 0 to 1, which is one minus pi by four. The final value is pi by eight minus half of one minus pi by four, which is pi by four minus one half.

Q2 (CBSE Board 2026, 3 marks) Find the integral of x plus two over x minus two, with respect to x. Answer: The division step is the key: once the integrand is written as a constant plus a simple fraction, each piece is a standard form, and no substitution or partial fraction is needed. Divide first, because the numerator and denominator have the same degree. Write x plus two over x minus two as one plus four over x minus two, since x plus two equals (x minus two) plus four. The integral of one is x and the integral of four over x minus two is four times the logarithm of the absolute value of x minus two. The answer is x plus four log of the absolute value of x minus two, plus C. Differentiating returns one plus four over x minus two, which is the original integrand.

Q3 (CBSE Board 2026, 3 marks) If I1 is the integral from minus pi by four to pi by four of dx over one plus cosine two x, and I2 is the integral from minus one half to one half of the absolute value of x, show that I1 minus four I2 equals zero. Answer: In I1, one plus cosine two x is two cosine squared x, so the integrand is half secant squared x. An antiderivative is half tangent x, and between the limits the value is half of tangent of pi by four minus tangent of minus pi by four, which is half of one minus minus one, equal to one. For I2 the integrand is even, so the integral is twice the integral from 0 to one half of x, which is twice of x squared over two at one half, equal to one quarter. Therefore I1 minus four I2 is one minus one, which is zero.

Q4 (CBSE Board 2026, 3 marks) Evaluate the integral from 0 to 1 of the logarithm of one plus x squared, with respect to x. Answer: The boundary term at the upper limit is log two because one plus one squared is two and the logarithm of one is zero at the lower limit. Keeping that term separate from the remaining integral avoids sign errors. Use by parts with u equal to the logarithm of one plus x squared and v prime equal to one, so v is x. The integral equals x log of one plus x squared from 0 to 1 minus the integral from 0 to 1 of two x squared over one plus x squared. The boundary term is log two. Writing two x squared over one plus x squared as two minus two over one plus x squared, the remaining integral is two x minus two inverse tangent x from 0 to 1, which is two minus pi by two. The answer is log two minus two plus pi by two.

Q5 (CBSE Board 2026, 3 marks) Evaluate the integral from 0 to pi by two of sine to the power 2026 x divided by the sum of sine to the power 2026 x and cosine to the power 2026 x, with respect to x. Answer: Let the integral be I and replace x by pi by two minus x. Sine of pi by two minus x is cosine x, so the transformed integral has the two powers exchanged. Adding the two expressions for I makes the numerator equal to the denominator, so twice I is the integral from 0 to pi by two of one, which is pi by two. Therefore I is pi by four, independent of the exponent as long as the two powers are equal.

Q6 (CBSE Board 2026, 3 marks) Find the integral of x plus sine x over one plus cosine x, with respect to x. Answer: Split the integrand. For x over one plus cosine x, write the denominator as two cosine squared of x by two and integrate by parts, giving x tangent of x by two plus two log of the absolute value of cosine of x by two. For sine x over one plus cosine x, the numerator is the negative derivative of the denominator, so the integral is minus the logarithm of one plus cosine x, which equals minus log two minus twice the log of cosine of x by two. The logarithmic terms cancel and minus log two is absorbed into the constant, so the answer is x tangent of x by two plus C.

Q7 (CBSE Board 2025, 3 marks) Evaluate the integral of the square root of four x squared minus four x plus ten, with respect to x. Answer: Complete the square inside the root: four x squared minus four x plus ten is two x minus one whole squared plus nine. Put t equal to two x minus one, so that dt is two dx and dx is dt by two. The integral becomes half the integral of the square root of t squared plus nine dt. The standard result for that integral is half t times the root plus nine over two times the logarithm of the absolute value of t plus the root. Multiplying by the half already outside gives one quarter of two x minus one times the root of four x squared minus four x plus ten, plus nine quarters of the logarithm of the absolute value of two x minus one plus the root of four x squared minus four x plus ten, plus C. Differentiating this answer returns the original square root, which confirms the coefficients.

Q8 (CBSE Board 2024, 2 marks) Evaluate the integral from 0 to 4 of the absolute value of x minus 2, with respect to x. Answer: Reading the region as two triangles is the fastest check on the answer: each triangle has base two and height two, so each contributes two square units, and the two pieces add to four. Split the interval at x equal to 2. From 0 to 2 the absolute value is 2 minus x, and its integral is the area of a triangle of base two and height two, which is two. From 2 to 4 the absolute value is x minus two, and its integral is again two. Adding the two pieces gives four square units.

Q9 (CBSE Board 2023, 3 marks) Evaluate the integral from 0 to pi by two of the difference between the logarithm of sine x and the logarithm of two cosine x, with respect to x. Answer: Write the integral as the integral of log sine x minus the integral of log cosine x minus the integral of log two. The first two integrals are each equal to minus pi by two times log two, a standard result obtained with the reflective property, and the third integral is pi by two times log two. Therefore the value is minus pi by two log two plus pi by two log two minus pi by two log two, which simplifies to minus pi by two log two.

Q10 (CBSE Board 2023, 3 marks) Evaluate the integral from 0 to pi by two of the absolute value of sine x minus cosine x, with respect to x. Answer: The expression inside the absolute value changes sign at pi by four. On the first piece the absolute value is cosine x minus sine x, whose integral is sine x plus cosine x evaluated from 0 to pi by four, giving root two minus one. On the second piece the absolute value is sine x minus cosine x, whose integral is minus cosine x minus sine x from pi by four to pi by two, again root two minus one. Adding the two pieces gives twice root two minus two.