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} = \dfrac12(\log2-\log1).
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 \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.
Substituting x=\tan\theta gives \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=2\tan^{-1}x for |x|\le1, reducing the integral to 2\displaystyle\int_0^1 \tan^{-1}x\,dx.
This equals 2\left[x\tan^{-1}x-\dfrac12\log(1+x^2)\right]_0^1.
Put x+2=t^2, so x=t^2-2, dx=2t\,dt; the limits become t=\sqrt2 to t=2.
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}{15}. At t=\sqrt2: -\dfrac{16\sqrt2}{15}.
Put t=\cos x, so dt=-\sin x\,dx; the limits become t=1 to t=0. The integral becomes \displaystyle\int_0^1 \dfrac{dt}{1+t^2} = \tan^{-1}1-\tan^{-1}0.
Complete the square: x+4-x^2 = \dfrac{17}{4}-\left(x-\dfrac12\right)^2.
Put t=x-\dfrac12 and apply \displaystyle\int \dfrac{dt}{a^2-t^2} = \dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right| with a=\dfrac{\sqrt{17}}{2}.
Evaluating at x=2 and x=0 and combining the logarithms.
Complete the square: x^2+2x+5=(x+1)^2+4. Anti derivative: \dfrac12\tan^{-1}\left(\dfrac{x+1}{2}\right).
At x=1: \dfrac{\pi}{8}. At x=-1: 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 anti derivative is \dfrac{e^{2x}}{2x}.
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}, and the integrand becomes \left(\dfrac{1}{x^2}-1\right)^{1/3}\dfrac{1}{x^3}.
Put t=\dfrac{1}{x^2}-1, so \dfrac{dx}{x^3}=-\dfrac{dt}{2}; the limits become t=8 at x=\dfrac13 and t=0 at x=1.
The integral becomes \dfrac12\displaystyle\int_0^8 t^{1/3}\,dt = \dfrac38\left(8^{4/3}\right) = \dfrac38\times16.
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.
Expert CBSE Coaching · Class 9–12