Where Mindful Learning Begins
Where Mindful Learning Begins
Chapter 7 Ex 7.9 covers evaluating definite integrals by substitution — all 10 questions solved step by step, changing the limits of integration to the new variable so there is no need to substitute back before plugging in the limits.
The one habit this exercise builds is easy to state and easy to forget under exam pressure: the moment you substitute t for x in a definite integral, convert the limits to t as well, and finish the entire calculation in the new variable. Skipping this and substituting back to x at the end works too, but it's slower and adds an unnecessary place to make an error. Most questions here reuse substitution and standard-form techniques from earlier in the chapter — the new piece is just carrying the limits through correctly. Q9 and Q10 are MCQs, with Q10 testing the Fundamental Theorem of Calculus directly rather than substitution.
Put t=x^2+1, so dt=2x\,dx; the limits become t=1 to t=2. The integral becomes \dfrac12\displaystyle\int_1^2 \dfrac{dt}{t}.
Integrating: \dfrac12\Big[\log|t|\Big]_1^2 = \dfrac12(\log2-\log1) = \dfrac12(\log2-0).
Write \cos^5\varphi = (1-\sin^2\varphi)^2\cos\varphi. Put t=\sin\varphi, so dt=\cos\varphi\,d\varphi; the limits become t=0 to t=1.
The integral becomes \displaystyle\int_0^1 t^{1/2}(1-t^2)^2\,dt. Expanding (1-t^2)^2=1-2t^2+t^4 and multiplying by t^{1/2}:
\displaystyle\int_0^1 \left(t^{1/2}-2t^{5/2}+t^{9/2}\right)dt = \left[\dfrac23 t^{3/2}-\dfrac47 t^{7/2}+\dfrac{2}{11}t^{11/2}\right]_0^1
Putting in the limits: \dfrac23-\dfrac47+\dfrac{2}{11}=\dfrac{154-132+42}{231}.
Put x=\tan\theta. Then \dfrac{2x}{1+x^2}=\dfrac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta, so \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=\sin^{-1}(\sin2\theta)=2\theta=2\tan^{-1}x, valid since x\in[0,1] gives 2\theta\in\left[0,\dfrac{\pi}{2}\right].
The integral reduces to 2\displaystyle\int_0^1 \tan^{-1}x\,dx. Integrating by parts (taking 1 as the second function):
\displaystyle\int \tan^{-1}x\,dx = x\tan^{-1}x-\int\dfrac{x}{1+x^2}\,dx = x\tan^{-1}x-\dfrac12\log(1+x^2)
So the integral equals 2\left[x\tan^{-1}x-\dfrac12\log(1+x^2)\right]_0^1 = 2\left[\left(\dfrac{\pi}{4}-\dfrac12\log2\right)-0\right].
Put x+2=t^2, so x=t^2-2, dx=2t\,dt; the limits become t=\sqrt2 to t=2.
Since x\sqrt{x+2}\,dx=(t^2-2)\cdot t\cdot 2t\,dt=2(t^4-2t^2)\,dt, the integral becomes 2\displaystyle\int_{\sqrt2}^2 (t^4-2t^2)\,dt = 2\left[\dfrac{t^5}{5}-\dfrac{2t^3}{3}\right]_{\sqrt2}^2.
At t=2: \dfrac{32}{5}-\dfrac{16}{3}=\dfrac{16}{15}. At t=\sqrt2: \dfrac{4\sqrt2}{5}-\dfrac{4\sqrt2}{3}=-\dfrac{8\sqrt2}{15}. Hence the value is 2\left[\dfrac{16}{15}+\dfrac{8\sqrt2}{15}\right].
Put t=\cos x, so dt=-\sin x\,dx; the limits become t=1 to t=0. The integral becomes -\displaystyle\int_1^0 \dfrac{dt}{1+t^2}. Reversing the limits removes the minus sign: \displaystyle\int_0^1 \dfrac{dt}{1+t^2}.
Integrating: \Big[\tan^{-1}t\Big]_0^1 = \tan^{-1}1-\tan^{-1}0 = \dfrac{\pi}{4}-0.
Complete the square: x+4-x^2 = -(x^2-x)+4 = -\left[\left(x-\dfrac12\right)^2-\dfrac14\right]+4 = \dfrac{17}{4}-\left(x-\dfrac12\right)^2.
Put t=x-\dfrac12, so dt=dx; the limits become t=-\dfrac12 to t=\dfrac32. With a=\dfrac{\sqrt{17}}{2} the integral is \displaystyle\int_{-1/2}^{3/2} \dfrac{dt}{a^2-t^2}.
Apply \displaystyle\int \dfrac{dt}{a^2-t^2} = \dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|, where \dfrac{1}{2a}=\dfrac{1}{\sqrt{17}}:
\dfrac{1}{\sqrt{17}}\left[\log\left|\dfrac{\sqrt{17}+2t}{\sqrt{17}-2t}\right|\right]_{-1/2}^{3/2} = \dfrac{1}{\sqrt{17}}\left[\log\dfrac{\sqrt{17}+3}{\sqrt{17}-3}-\log\dfrac{\sqrt{17}-1}{\sqrt{17}+1}\right]
Combine the logs and multiply out: \dfrac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}=\dfrac{20+4\sqrt{17}}{20-4\sqrt{17}}=\dfrac{5+\sqrt{17}}{5-\sqrt{17}}.
Complete the square: x^2+2x+5=(x+1)^2+4. So the integral is \displaystyle\int_{-1}^1 \dfrac{dx}{(x+1)^2+2^2}. Using \displaystyle\int \dfrac{dx}{x^2+a^2}=\dfrac1a\tan^{-1}\dfrac xa with x+1 in place of x and a=2, the antiderivative is \dfrac12\tan^{-1}\left(\dfrac{x+1}{2}\right).
At x=1: \dfrac12\tan^{-1}(1)=\dfrac12\cdot\dfrac{\pi}{4}=\dfrac{\pi}{8}. At x=-1: \dfrac12\tan^{-1}(0)=0.
Subtracting: \dfrac{\pi}{8}-0.
Try h(x)=\dfrac{1}{2x}, so h'(x)=-\dfrac{1}{2x^2}.
Then \dfrac{d}{dx}\left[e^{2x}h(x)\right] = e^{2x}[2h(x)+h'(x)] = e^{2x}\left[\dfrac{1}{x}-\dfrac{1}{2x^2}\right] — exactly the integrand, so the antiderivative is \dfrac{e^{2x}}{2x}.
Check: 2h(x)=\dfrac1x and h'(x)=-\dfrac{1}{2x^2}, which are the two terms of the bracket.
Putting in the limits: \left[\dfrac{e^{2x}}{2x}\right]_1^2=\dfrac{e^4}{4}-\dfrac{e^2}{2}.
Write x-x^3=x^3\left(\dfrac{1}{x^2}-1\right), so (x-x^3)^{1/3}=x\left(\dfrac{1}{x^2}-1\right)^{1/3}. Dividing by x^4, the integrand becomes \left(\dfrac{1}{x^2}-1\right)^{1/3}\dfrac{1}{x^3}.
Put t=\dfrac{1}{x^2}-1. Then dt=-\dfrac{2}{x^3}\,dx, so \dfrac{dx}{x^3}=-\dfrac{dt}{2}. The limits become t=9-1=8 at x=\dfrac13 and t=1-1=0 at x=1.
The integral becomes -\dfrac12\displaystyle\int_8^0 t^{1/3}\,dt = \dfrac12\displaystyle\int_0^8 t^{1/3}\,dt (reversing the limits removes the minus sign).
Integrating: \dfrac12\left[\dfrac{3}{4}t^{4/3}\right]_0^8 = \dfrac38\left(8^{4/3}\right). Since 8^{4/3}=\left(8^{1/3}\right)^4=2^4=16, this is \dfrac38\times16 = 6.
By the First Fundamental Theorem of Calculus, if f(x)=\displaystyle\int_0^x g(t)\,dt, then f'(x)=g(x) directly — no integration is needed.
Here g(t)=t\sin t, so replacing t by x gives f'(x)=x\sin x.
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