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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.1 1–10 (part 1 of 27)

  1. Find an anti derivative (or integral) of the following functions by the method of inspection.

    Exercise 1

    sin⁡2x\displaystyle \sin 2 x

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2} \cos 2 x\)
    By the method of inspection, look for a function whose derivative is \(\displaystyle \sin 2x\). The cosine is the natural candidate, and the chain rule supplies the inner factor: \[\frac{d}{dx}(\cos 2x) = -2\sin 2x.\] Divide through by \(\displaystyle -2\): \[\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x.\] The factor \(\displaystyle \frac{1}{2}\) is exactly what compensates for the \(\displaystyle 2\) the chain rule produces — that is the step to get right. \[\int \sin 2x\, dx = -\frac{1}{2}\cos 2x + C.\]

  2. Exercise 2

    cos⁡3x\displaystyle \cos 3 x

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    NCERT’s answer
    \(\displaystyle \frac{1}{3} \sin 3 x\)
    By inspection, differentiate the natural candidate \(\displaystyle \sin 3x\) using the chain rule: \[\frac{d}{dx}(\sin 3x) = 3\cos 3x.\] Divide by \(\displaystyle 3\): \[\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right) = \cos 3x.\] So an anti derivative of \(\displaystyle \cos 3x\) is \(\displaystyle \frac{1}{3}\sin 3x\), and \[\int \cos 3x\, dx = \frac{1}{3}\sin 3x + C.\]

  3. Exercise 3

    e2x\displaystyle e^{2 x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{2} e^{2 x}\)
    By inspection, the exponential reproduces itself on differentiation, with the chain rule contributing the exponent's coefficient: \[\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}.\] Hence \[\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x},\] so \[\int e^{2x}\, dx = \frac{1}{2}e^{2x} + C.\]

  4. Exercise 4

    (ax+b)2\displaystyle (a x+b)^{2}

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    NCERT’s answer
    \(\displaystyle \frac{1}{3 a}(a x+b)^{3}\)
    Take \(\displaystyle a \neq 0\). By inspection, try the next power, \(\displaystyle (ax+b)^{3}\), and differentiate by the chain rule: \[\frac{d}{dx}(ax+b)^{3} = 3(ax+b)^{2}\cdot a = 3a\,(ax+b)^{2}.\] Divide by \(\displaystyle 3a\): \[\frac{d}{dx}\left(\frac{(ax+b)^{3}}{3a}\right) = (ax+b)^{2}.\] The \(\displaystyle \frac{1}{a}\) is the factor most often dropped here: it is the derivative of the inner function \(\displaystyle ax+b\) being undone. \[\int (ax+b)^{2}\, dx = \frac{(ax+b)^{3}}{3a} + C \qquad (a \neq 0).\] If \(\displaystyle a = 0\) the function is the constant \(\displaystyle b^{2}\), whose anti derivative is \(\displaystyle b^{2}x + C\).

  5. Exercise 5

    sin⁡2x−4e3x\displaystyle \sin 2 x-4 e^{3 x}

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    NCERT’s answer
    \(\displaystyle -\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}\)
    The anti derivative is linear, so treat the two terms separately. For the first term, as in Question $\displaystyle 1$, \(\displaystyle \frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x\). For the second, \(\displaystyle \frac{d}{dx}\left(e^{3x}\right) = 3e^{3x}\), so \(\displaystyle \frac{d}{dx}\left(\frac{1}{3}e^{3x}\right) = e^{3x}\) and therefore \[\frac{d}{dx}\left(-\frac{4}{3}e^{3x}\right) = -4e^{3x}.\] Adding, \[\frac{d}{dx}\left(-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}\right) = \sin 2x - 4e^{3x}.\] Hence \[\int \left(\sin 2x - 4e^{3x}\right) dx = -\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x} + C.\]
  6. Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:

    Exercise 6

    ∫(4e3x+1)dx\displaystyle \int\left(4 e^{3 x}+1\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{4}{3} e^{3 x}+x+\mathrm{C}\)
    Use the linearity of the integral, \(\displaystyle \int (f + g)\,dx = \int f\,dx + \int g\,dx\), together with \(\displaystyle \int e^{kx}dx = \frac{1}{k}e^{kx}\) (\(\displaystyle k \neq 0\)) and \(\displaystyle \int 1\,dx = x\): \[\int \left(4e^{3x} + 1\right) dx = 4\int e^{3x}\,dx + \int 1\,dx = 4\cdot \frac{e^{3x}}{3} + x + C.\] So \[\int \left(4e^{3x} + 1\right) dx = \frac{4}{3}e^{3x} + x + C.\]

  7. Exercise 7

    ∫x2(1−1x2)dx\displaystyle \int x^{2}\left(1-\frac{1}{x^{2}}\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{x^{3}}{3}-x+\mathrm{C}\)
    There is no product rule for integration, so simplify the integrand first (valid for \(\displaystyle x \neq 0\)): \[x^{2}\left(1 - \frac{1}{x^{2}}\right) = x^{2} - 1.\] Now apply the power rule \(\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}\ (n \neq -1)\): \[\int \left(x^{2} - 1\right) dx = \frac{x^{3}}{3} - x + C.\] Hence \[\int x^{2}\left(1 - \frac{1}{x^{2}}\right) dx = \frac{x^{3}}{3} - x + C, \qquad x \neq 0.\]

  8. Exercise 8

    ∫(ax2+bx+c)dx\displaystyle \int\left(a x^{2}+b x+c\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+\mathrm{C}\)
    By linearity, integrate term by term with the power rule \(\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}\) (here \(\displaystyle a, b, c\) are constants and come out of the integrals): \[\int \left(ax^{2} + bx + c\right) dx = a\int x^{2}dx + b\int x\,dx + c\int 1\,dx.\] So \[\int \left(ax^{2} + bx + c\right) dx = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + cx + C.\]

  9. Exercise 9

    ∫(2x2+ex)dx\displaystyle \int\left(2 x^{2}+e^{x}\right) d x

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    NCERT’s answer
    \(\displaystyle \frac{2}{3} x^{3}+e^{x}+\mathrm{C}\)
    By linearity, using the power rule for the first term and \(\displaystyle \int e^{x}dx = e^{x}\) for the second: \[\int \left(2x^{2} + e^{x}\right) dx = 2\int x^{2}dx + \int e^{x}dx = 2\cdot\frac{x^{3}}{3} + e^{x} + C.\] Hence \[\int \left(2x^{2} + e^{x}\right) dx = \frac{2}{3}x^{3} + e^{x} + C.\]

  10. Exercise 10

    ∫(x−1x)2dx\displaystyle \int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x

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    NCERT’s answer
    \(\displaystyle \frac{x^{2}}{2}+\log |x|-2 x+\mathrm{C}\)
    Expand the square before integrating — the middle term is where the marks are lost: \[\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} = (\sqrt{x})^{2} - 2\cdot\sqrt{x}\cdot\frac{1}{\sqrt{x}} + \left(\frac{1}{\sqrt{x}}\right)^{2} = x - 2 + \frac{1}{x}.\] The presence of \(\displaystyle \sqrt{x}\) and \(\displaystyle \frac{1}{\sqrt{x}}\) restricts the domain to \(\displaystyle x > 0\). Now integrate term by term, using \(\displaystyle \int \frac{1}{x}dx = \log|x|\): \[\int \left(x - 2 + \frac{1}{x}\right) dx = \frac{x^{2}}{2} - 2x + \log x + C.\] So \[\int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} dx = \frac{x^{2}}{2} - 2x + \log x + C, \qquad x > 0.\]