NCERT Class 12 Mathematics — Integrals, Exercise 7.9. All 10 questions solved.

Exercise 7.8 evaluated definite integrals whose antiderivatives you already knew. This one adds substitution, and with it the single point the whole exercise is built around: when the variable changes, the limits change with it.

There are two legitimate ways to finish a substituted definite integral. Either resubstitute back to $x$ and use the original limits, or — much cleaner — convert the limits to the new variable and never go back:

$$\int_a^b f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt$$

Key insight. Changing the limits is not a shortcut, it is the point. Once the integral is written in $t$ with $t$-limits, the problem is finished — no resubstitution, no back-conversion, and no chance of evaluating a $t$-expression at an $x$-value, which is the commonest way this exercise goes wrong.

Note also that the new limits need not be in increasing order. In question 5 the substitution $t = \cos x$ sends $[0, \tfrac{\pi}{2}]$ to $[1, 0]$, and the minus sign from $dt = -\sin x\,dx$ is exactly what flips them back.

Evaluate the integrals in Exercises 1 to 8 using substitution.

Question 1

$\displaystyle\int_{0}^{1} \frac{x}{x^2 + 1}\,dx$

Solution. Put $t = x^2 + 1$, so $dt = 2x\,dx$. The limits move with the substitution: $x = 0$ gives $t = 1$, and $x = 1$ gives $t = 2$.

$$\frac12\int_1^2 \frac{dt}{t} = \frac12\Big[\log t\Big]_1^2 = \frac12\log 2$$

Question 2

$\displaystyle\int_{0}^{\pi/2} \sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

Solution. The odd power of cosine is the signal. Peel off one $\cos\phi$ for the differential and convert the rest with $\cos^2\phi = 1 – \sin^2\phi$:

$$\cos^5\phi\,d\phi = \left(1 – \sin^2\phi\right)^2 \cos\phi\,d\phi$$

Put $t = \sin\phi$, $dt = \cos\phi\,d\phi$, with limits $0$ to $1$:

$$\int_0^1 \sqrt{t}\,(1 – t^2)^2\,dt = \int_0^1 \left(t^{1/2} – 2t^{5/2} + t^{9/2}\right) dt$$

$$= \frac23 – \frac47 + \frac{2}{11} = \frac{154 – 132 + 42}{231}$$

Question 3

$\displaystyle\int_{0}^{1} \sin^{-1}\left(\frac{2x}{1 + x^2}\right) dx$

Solution. The argument is the double-angle form for sine. Put $x = \tan\theta$, so that

$$\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta$$

For $x \in [0, 1]$ we have $\theta \in \left[0, \tfrac{\pi}{4}\right]$, so $2\theta \in \left[0, \tfrac{\pi}{2}\right]$ and $\sin^{-1}(\sin 2\theta) = 2\theta$ without any adjustment. With $dx = \sec^2\theta\,d\theta$:

$$\int_0^{\pi/4} 2\theta\sec^2\theta\,d\theta$$

Integrate by parts, taking $2\theta$ as the first function:

$$= \Big[2\theta\tan\theta\Big]_0^{\pi/4} – 2\int_0^{\pi/4}\tan\theta\,d\theta = \frac{\pi}{2} – 2\Big[\log\sec\theta\Big]_0^{\pi/4}$$

Since $\sec\tfrac{\pi}{4} = \sqrt2$, the second term is $2\log\sqrt2 = \log 2$.

Question 4

$\displaystyle\int_{0}^{2} x\sqrt{x + 2}\,dx$  (Put $x + 2 = t^2$)

Solution. NCERT suggests $x + 2 = t^2$ rather than $x + 2 = t$, because squaring clears the root entirely instead of leaving a half-power. Then $dx = 2t\,dt$ and $x = t^2 – 2$. The limits: $x = 0$ gives $t = \sqrt2$, and $x = 2$ gives $t = 2$.

$$\int_{\sqrt2}^{2} (t^2 – 2)\cdot t \cdot 2t\,dt = 2\int_{\sqrt2}^{2}\left(t^4 – 2t^2\right) dt = 2\left[\frac{t^5}{5} – \frac{2t^3}{3}\right]_{\sqrt2}^{2}$$

At $t = 2$ the bracket is $\tfrac{32}{5} – \tfrac{16}{3} = \tfrac{16}{15}$. At $t = \sqrt2$ it is $\tfrac{4\sqrt2}{5} – \tfrac{4\sqrt2}{3} = -\tfrac{8\sqrt2}{15}$. Doubling the difference gives $\tfrac{32 + 16\sqrt2}{15}$, which factors as:

Question 5

$\displaystyle\int_{0}^{\pi/2} \frac{\sin x}{1 + \cos^2 x}\,dx$

Solution. Put $t = \cos x$, so $dt = -\sin x\,dx$. The limits invert: $x = 0$ gives $t = 1$, and $x = \tfrac{\pi}{2}$ gives $t = 0$.

$$-\int_1^0 \frac{dt}{1+t^2} = \int_0^1 \frac{dt}{1+t^2} = \Big[\tan^{-1}t\Big]_0^1$$

The minus sign and the reversed limits cancel each other — this is worth doing explicitly rather than by inspection.

Question 6

$\displaystyle\int_{0}^{2} \frac{dx}{x + 4 – x^2}$

Solution. Complete the square in the denominator, keeping the sign in front of $x^2$ in mind:

$$x + 4 – x^2 = -\left(x^2 – x – 4\right) = \frac{17}{4} – \left(x – \frac12\right)^2$$

This is the $a^2 – t^2$ form with $a = \tfrac{\sqrt{17}}{2}$ and $t = x – \tfrac12$:

$$\int\frac{dt}{a^2 – t^2} = \frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|$$

Here $\tfrac{1}{2a} = \tfrac{1}{\sqrt{17}}$. Evaluating between $x = 0$ and $x = 2$:

$$\frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17}+3}{\sqrt{17}-3} – \log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]$$

Rationalising both ratios and combining gives $\tfrac{21 + 5\sqrt{17}}{4}$ inside a single logarithm.

Question 7

$\displaystyle\int_{-1}^{1} \frac{dx}{x^2 + 2x + 5}$

Solution. Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$. Put $t = x+1$, which shifts the limits from $[-1, 1]$ to $[0, 2]$:

$$\int_0^2 \frac{dt}{t^2 + 2^2} = \frac12\Big[\tan^{-1}\frac{t}{2}\Big]_0^2 = \frac12\left(\frac{\pi}{4} – 0\right)$$

A shift is still a substitution, and the limits still have to move.

Question 8

$\displaystyle\int_{1}^{2} \left(\frac{1}{x} – \frac{1}{2x^2}\right) e^{2x}\,dx$

Solution. This is the $\int e^{ax}\left[a f(x) + f'(x)\right] dx = e^{ax}f(x)$ pattern with $a = 2$. The whole question is spotting $f$. Try $f(x) = \dfrac{1}{2x}$:

$$f'(x) = -\frac{1}{2x^2}, \qquad 2f(x) + f'(x) = \frac1x – \frac{1}{2x^2}$$

which is exactly the bracket. So the antiderivative is $\dfrac{e^{2x}}{2x}$:

$$\left[\frac{e^{2x}}{2x}\right]_1^2 = \frac{e^4}{4} – \frac{e^2}{2}$$

Choose the correct answer in Exercises 9 and 10.

Question 9

The value of the integral $\displaystyle\int_{1/3}^{1} \frac{\left(x – x^3\right)^{1/3}}{x^4}\,dx$ is

Solution. Factor $x^3$ out of the cube root rather than $x$ — that is what makes the substitution appear:

$$x – x^3 = x^3\left(\frac{1}{x^2} – 1\right), \qquad \left(x – x^3\right)^{1/3} = x\left(\frac{1}{x^2} – 1\right)^{1/3}$$

The integrand becomes $\dfrac{1}{x^3}\left(\dfrac{1}{x^2} – 1\right)^{1/3}$. Now put $t = \dfrac{1}{x^2} – 1$, so $dt = -\dfrac{2}{x^3}\,dx$. The limits: $x = \tfrac13$ gives $t = 8$, and $x = 1$ gives $t = 0$.

$$-\frac12\int_8^0 t^{1/3}\,dt = \frac12\int_0^8 t^{1/3}\,dt = \frac12 \cdot \frac34\Big[t^{4/3}\Big]_0^8 = \frac38 \cdot 16 = 6$$

A note on the printed answer. The NCERT answer key gives (D) $4$ here. That is a misprint. Both the substitution above and a direct numerical evaluation of the integral give $6$, so the correct option is (A). If your work led you to $6$, your work was right.

Question 10

If $f(x) = \displaystyle\int_{0}^{x} t\sin t\,dt$, then $f'(x)$ is

Solution. No integration is needed. The first Fundamental Theorem of Calculus says that if $f(x) = \int_a^x g(t)\,dt$ then $f'(x) = g(x)$ — differentiation undoes the integral directly. Here $g(t) = t\sin t$, so

$$f'(x) = x\sin x$$

Option (A) is what you get by actually evaluating the integral by parts and then differentiating carelessly; options (A) and (D) are both derivatives of the antiderivative expression rather than of $f$.

Common mistakes

• Keeping the old limits after substituting. If you write the integral in $t$, the limits must be $t$-values. Evaluating a $t$-expression at $x = 0$ and $x = 1$ is the single biggest source of wrong answers in this exercise.
• Resubstituting and changing the limits. Do one or the other. Doing both applies the change twice.
• Panicking when the new limits come out reversed. In questions 5 and 9 the upper limit is smaller than the lower one. That is fine — the minus sign from $dt$ flips them.
• Substituting $x + 2 = t$ in question 4. It works, but leaves $(t-2)\sqrt{t}$, i.e. half-powers. NCERT’s $x + 2 = t^2$ turns the whole integrand into a polynomial.
• Factoring $x$ instead of $x^3$ in question 9. Pulling out $x$ gives $(1 – x^2)^{1/3}$, which leads nowhere. Pulling out $x^3$ produces $\left(\tfrac{1}{x^2} – 1\right)^{1/3}$, whose derivative is already sitting in the $\tfrac{1}{x^3}$ outside.
• Integrating in question 10. The Fundamental Theorem answers it in one line. Evaluating $\int_0^x t\sin t\,dt$ first and then differentiating is three times the work and invites a sign error.

Practise next

• Exercise 7.10 — the properties of definite integrals, which handle several integrals that substitution alone struggles with.
• Exercise 7.6 — integration by parts, used in questions 3 and 8 here.