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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.2 1–10 (part 3 of 27)

  1. Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:

    Exercise 1

    2x1+x2\displaystyle \frac{2 x}{1+x^{2}}

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    NCERT’s answer
    \(\displaystyle \log \left(1+x^{2}\right)+\mathrm{C}\)
    Use the method of substitution: if \(\displaystyle t=g(x)\), then \(\displaystyle g'(x)\,dx=dt\). Here the numerator is exactly the derivative of the denominator, so put \[t=1+x^{2}\quad\Rightarrow\quad dt=2x\,dx.\] Then \[\int\frac{2x}{1+x^{2}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Since \(\displaystyle 1+x^{2}>0\) for every real \(\displaystyle x\), the modulus may be dropped. Resubstituting \(\displaystyle t=1+x^{2}\), \[\int\frac{2x}{1+x^{2}}\,dx=\log\left(1+x^{2}\right)+\mathrm{C}.\]

  2. Exercise 2

    (log⁡x)2x\displaystyle \frac{(\log x)^{2}}{x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}(\log |x|)^{3}+\mathrm{C}\)
    The factor \(\displaystyle \dfrac{1}{x}\) is the derivative of \(\displaystyle \log x\), which invites the substitution \[t=\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{(\log x)^{2}}{x}\,dx=\int t^{2}\,dt=\frac{t^{3}}{3}+\mathrm{C}.\] Resubstituting \(\displaystyle t=\log x\), \[\int\frac{(\log x)^{2}}{x}\,dx=\frac{(\log x)^{3}}{3}+\mathrm{C}.\]

  3. Exercise 3

    1x+xlog⁡x\displaystyle \frac{1}{x+x \log x}

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    NCERT’s answer
    \(\displaystyle \log |1+\log x|+C\)
    First take \(\displaystyle x\) common in the denominator — this is the step that makes the substitution visible: \[\frac{1}{x+x\log x}=\frac{1}{x\,(1+\log x)}.\] Put \[t=1+\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{dx}{x(1+\log x)}=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Hence \[\int\frac{1}{x+x\log x}\,dx=\log\left|1+\log x\right|+\mathrm{C},\qquad x>0,\ \log x\neq-1.\]

  4. Exercise 4

    sin⁡xsin⁡(cos⁡x)\displaystyle \sin x \sin (\cos x)

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    NCERT’s answer
    \(\displaystyle \cos (\cos x)+\mathrm{C}\)
    The outer function has argument \(\displaystyle \cos x\), and \(\displaystyle \sin x\) is (up to sign) its derivative. Put \[t=\cos x\quad\Rightarrow\quad dt=-\sin x\,dx,\ \text{ i.e. }\ \sin x\,dx=-dt.\] Watch the sign here — it is the usual slip. Then \[\int\sin x\,\sin(\cos x)\,dx=\int \sin t\,(-dt)=-\int\sin t\,dt=\cos t+\mathrm{C}.\] Resubstituting \(\displaystyle t=\cos x\), \[\int\sin x\,\sin(\cos x)\,dx=\cos(\cos x)+\mathrm{C}.\]

  5. Exercise 5

    sin⁡(ax+b)cos⁡(ax+b)\displaystyle \sin (a x+b) \cos (a x+b)

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    NCERT’s answer
    \(\displaystyle -\frac{1}{4 a} \cos 2(a x+b)+\mathrm{C}\)
    Use the identity \(\displaystyle 2\sin\theta\cos\theta=\sin 2\theta\) with \(\displaystyle \theta=ax+b\): \[\sin(ax+b)\cos(ax+b)=\frac{1}{2}\sin\bigl(2(ax+b)\bigr).\] Therefore, using \(\displaystyle \int\sin(kx+c)\,dx=-\dfrac{\cos(kx+c)}{k}\) with \(\displaystyle k=2a\ (a\neq0)\), \[\int\sin(ax+b)\cos(ax+b)\,dx=\frac{1}{2}\int\sin(2ax+2b)\,dx=\frac{1}{2}\left(-\frac{\cos(2ax+2b)}{2a}\right)+\mathrm{C}.\] Hence \[\int\sin(ax+b)\cos(ax+b)\,dx=-\frac{\cos 2(ax+b)}{4a}+\mathrm{C}.\] (The substitution \(\displaystyle t=\sin(ax+b)\) gives \(\displaystyle \dfrac{\sin^{2}(ax+b)}{2a}+\mathrm{C}\); the two answers differ only by the constant \(\displaystyle \dfrac{1}{4a}\).)

  6. Exercise 6

    ax+b\displaystyle \sqrt{a x+b}

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    NCERT’s answer
    \(\displaystyle \frac{2}{3 a}(a x+b)^{\frac{3}{2}}+\mathrm{C}\)
    Put \[t=ax+b\quad\Rightarrow\quad dt=a\,dx,\ \text{ i.e. }\ dx=\frac{dt}{a}\qquad(a\neq0).\] Then, by the power rule \(\displaystyle \int t^{n}\,dt=\dfrac{t^{n+1}}{n+1}\) with \(\displaystyle n=\tfrac12\), \[\int\sqrt{ax+b}\,dx=\frac{1}{a}\int t^{1/2}\,dt=\frac{1}{a}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int\sqrt{ax+b}\,dx=\frac{2}{3a}(ax+b)^{3/2}+\mathrm{C}.\]

  7. Exercise 7

    xx+2\displaystyle x \sqrt{x+2}

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    NCERT’s answer
    \(\displaystyle \frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+\mathrm{C}\)
    The factor \(\displaystyle x\) outside is not the derivative of \(\displaystyle x+2\), so substitute for the bracket and re-express \(\displaystyle x\) in the new variable — that back-substitution is the step to get right. Put \[t=x+2\quad\Rightarrow\quad x=t-2,\quad dx=dt.\] Then \[\int x\sqrt{x+2}\,dx=\int (t-2)\,t^{1/2}\,dt=\int\left(t^{3/2}-2t^{1/2}\right)dt=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+\mathrm{C}.\] Resubstituting \(\displaystyle t=x+2\), \[\int x\sqrt{x+2}\,dx=\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+\mathrm{C}.\]

  8. Exercise 8

    x1+2x2\displaystyle x \sqrt{1+2 x^{2}}

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    NCERT’s answer
    \(\displaystyle \frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+\mathrm{C}\)
    Here \(\displaystyle x\,dx\) is proportional to the differential of \(\displaystyle 1+2x^{2}\). Put \[t=1+2x^{2}\quad\Rightarrow\quad dt=4x\,dx,\ \text{ i.e. }\ x\,dx=\frac{dt}{4}.\] Then \[\int x\sqrt{1+2x^{2}}\,dx=\frac{1}{4}\int t^{1/2}\,dt=\frac{1}{4}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}=\frac{1}{6}t^{3/2}+\mathrm{C}.\] Hence \[\int x\sqrt{1+2x^{2}}\,dx=\frac{1}{6}\left(1+2x^{2}\right)^{3/2}+\mathrm{C}.\]

  9. Exercise 9

    (4x+2)x2+x+1\displaystyle (4 x+2) \sqrt{x^{2}+x+1}

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    NCERT’s answer
    \(\displaystyle \frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\mathrm{C}\)
    Notice that \(\displaystyle 4x+2=2(2x+1)\) and \(\displaystyle 2x+1\) is the derivative of \(\displaystyle x^{2}+x+1\). Put \[t=x^{2}+x+1\quad\Rightarrow\quad dt=(2x+1)\,dx,\ \text{ so }\ (4x+2)\,dx=2\,dt.\] Then \[\int(4x+2)\sqrt{x^{2}+x+1}\,dx=2\int t^{1/2}\,dt=2\cdot\frac{2}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int(4x+2)\sqrt{x^{2}+x+1}\,dx=\frac{4}{3}\left(x^{2}+x+1\right)^{3/2}+\mathrm{C}.\]

  10. Exercise 10

    1x−x\displaystyle \frac{1}{x-\sqrt{x}}

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    NCERT’s answer
    \(\displaystyle 2 \log |\sqrt{x}-1|+\mathrm{C}\)
    A square root of \(\displaystyle x\) with \(\displaystyle x\) itself present: substitute for the root. Put \[t=\sqrt{x}\quad\Rightarrow\quad x=t^{2},\quad dx=2t\,dt\qquad(x>0).\] Then \(\displaystyle x-\sqrt{x}=t^{2}-t=t(t-1)\), and the factor \(\displaystyle t\) cancels: \[\int\frac{dx}{x-\sqrt{x}}=\int\frac{2t\,dt}{t(t-1)}=2\int\frac{dt}{t-1}=2\log|t-1|+\mathrm{C}.\] Resubstituting \(\displaystyle t=\sqrt{x}\), \[\int\frac{1}{x-\sqrt{x}}\,dx=2\log\left|\sqrt{x}-1\right|+\mathrm{C},\qquad x>0,\ x\neq1.\]