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NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 1–10 (part 1 of 6)

  1. Verify the following :

    Exercise 1

    ∫2x−12x+3dx=x−log⁡∣(2x+3)2∣+C\displaystyle \int \frac{2 x-1}{2 x+3} d x=x-\log \left|(2 x+3)^2\right|+\mathrm{C}

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Differentiate the right-hand side.\[\frac{d}{dx}\Big[x-\log\left|(2x+3)^2\right|\Big]=\frac{d}{dx}\Big[x-2\log|2x+3|\Big] \] \[=1-2\cdot\frac{2}{2x+3} \] \[=\frac{2x+3-4}{2x+3}=\frac{2x-1}{2x+3} \]This is the integrand.Answer: Verified.

  2. Exercise 2

    ∫2x+3x2+3xdx=log⁡∣x2+3x∣+C\displaystyle \int \frac{2 x+3}{x^2+3 x} d x=\log \left|x^2+3 x\right|+\mathrm{C}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Differentiate the right-hand side.\[\frac{d}{dx}\log\left|x^2+3x\right|=\frac{1}{x^2+3x}\cdot\frac{d}{dx}\left(x^2+3x\right) \] \[=\frac{2x+3}{x^2+3x} \]This is the integrand.Answer: Verified.
  3. Evaluate the following:

    Exercise 3

    ∫(x2+2)dxx+1\displaystyle \int \frac{\left(x^2+2\right) d x}{x+1}

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    NCERT’s answer
    \(\displaystyle \frac{x^2}{2}-x+3 \log |x+1|+c\)
    Divide the numerator by \(\displaystyle x+1\).\[x^2+2=(x+1)(x-1)+3 \] \[\frac{x^2+2}{x+1}=x-1+\frac{3}{x+1} \] \[\int\frac{x^2+2}{x+1}\,dx=\int\left(x-1+\frac{3}{x+1}\right)dx \] \[=\frac{x^2}{2}-x+3\log|x+1|+C \]Answer: \(\displaystyle \dfrac{x^2}{2}-x+3\log|x+1|+C\)

  4. Exercise 4

    ∫e6log⁡x−e5log⁡xe4log⁡x−e3log⁡xdx\displaystyle \int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x

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    NCERT’s answer
    \(\displaystyle \frac{x^3}{3}+c\)
    For \(\displaystyle x>0\):\[e^{n\log x}=x^n \] \[\frac{e^{6\log x}-e^{5\log x}}{e^{4\log x}-e^{3\log x}}=\frac{x^6-x^5}{x^4-x^3} \] \[=\frac{x^5(x-1)}{x^3(x-1)}=x^2 \quad (x\neq 1) \] \[\int x^2\,dx=\frac{x^3}{3}+C \]Answer: \(\displaystyle \dfrac{x^3}{3}+C\)

  5. Exercise 5

    ∫(1+cos⁡x)x+sin⁡xdx\displaystyle \int \frac{(1+\cos x)}{x+\sin x} d x

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    NCERT’s answer
    \(\displaystyle \log |x+\sin x|+c\)
    Substitute the denominator.\[u=x+\sin x \;\Rightarrow\; du=(1+\cos x)\,dx \] \[\int\frac{1+\cos x}{x+\sin x}\,dx=\int\frac{du}{u} \] \[=\log|u|+C \] \[=\log|x+\sin x|+C \]Answer: \(\displaystyle \log|x+\sin x|+C\)

  6. Exercise 6

    ∫dx1+cos⁡x\displaystyle \int \frac{d x}{1+\cos x}

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    NCERT’s answer
    \(\displaystyle \tan \frac{x}{2}+\mathrm{C}\)
    Use the half-angle form.\[1+\cos x=2\cos^2\frac{x}{2} \] \[\int\frac{dx}{1+\cos x}=\frac12\int\sec^2\frac{x}{2}\,dx \] \[=\frac12\cdot 2\tan\frac{x}{2}+C \] \[=\tan\frac{x}{2}+C \]Answer: \(\displaystyle \tan\dfrac{x}{2}+C\)

  7. Exercise 7

    ∫tan⁡2xsec⁡4xdx\displaystyle \int \tan ^2 x \sec ^4 x d x

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    NCERT’s answer
    \(\displaystyle \frac{\tan ^5 x}{5}+\frac{\tan ^3 x}{3}+c\)
    Keep one \(\displaystyle \sec^2 x\) for the differential.\[\tan^2x\sec^4x=\tan^2x\,(1+\tan^2x)\sec^2x \] \[t=\tan x \;\Rightarrow\; dt=\sec^2x\,dx \] \[\int\left(t^2+t^4\right)dt=\frac{t^3}{3}+\frac{t^5}{5}+C \] \[=\frac{\tan^3x}{3}+\frac{\tan^5x}{5}+C \]Answer: \(\displaystyle \dfrac{\tan^3x}{3}+\dfrac{\tan^5x}{5}+C\)

  8. Exercise 8

    ∫sin⁡x+cos⁡x1+sin⁡2xdx\displaystyle \int \frac{\sin x+\cos x}{\sqrt{1+\sin 2 x}} d x

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    NCERT’s answer
    \(\displaystyle x+c\)
    Write the radicand as a square.\[1+\sin2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2 \] \[\sqrt{1+\sin2x}=\sin x+\cos x \quad (\sin x+\cos x>0) \] \[\int\frac{\sin x+\cos x}{\sin x+\cos x}\,dx=\int dx=x+C \]Where \(\displaystyle \sin x+\cos x<0\), the root is its negative and the result is \(\displaystyle -x+C\).Answer: \(\displaystyle x+C\)

  9. Exercise 9

    ∫1+sin⁡xdx\displaystyle \int \sqrt{1+\sin x} d x

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    NCERT’s answer
    \(\displaystyle -2 \cos \frac{x}{2}+2 \sin \frac{x}{2}+c\)
    Write the radicand as a perfect square.\[1+\sin x=\sin^2\tfrac{x}{2}+\cos^2\tfrac{x}{2}+2\sin\tfrac{x}{2}\cos\tfrac{x}{2}=\left(\sin\tfrac{x}{2}+\cos\tfrac{x}{2}\right)^2 \]\[\int\sqrt{1+\sin x}\,dx=\int\left(\sin\tfrac{x}{2}+\cos\tfrac{x}{2}\right)dx \]\[=-2\cos\tfrac{x}{2}+2\sin\tfrac{x}{2}+C \]Answer: \(\displaystyle 2\left(\sin\tfrac{x}{2}-\cos\tfrac{x}{2}\right)+C\)

  10. Exercise 10

    ∫xx+1dx\displaystyle \int \frac{x}{\sqrt{x}+1} d x (Hint : Put x=z\displaystyle \sqrt{x}=z)

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    NCERT’s answer
    \(\displaystyle 2\left[\frac{x \sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\log |\sqrt{x}+1|\right]+c\)
    Put \(\displaystyle \sqrt{x}=z\). \[x=z^2,\qquad dx=2z\,dz \] \[\int\frac{x}{\sqrt{x}+1}\,dx = \int\frac{2z^3}{z+1}\,dz \] \[z^3=(z+1)(z^2-z+1)-1 \] \[\frac{2z^3}{z+1}=2\left(z^2-z+1-\frac{1}{z+1}\right) \] \[\int\frac{2z^3}{z+1}\,dz = 2\left(\frac{z^3}{3}-\frac{z^2}{2}+z-\ln|z+1|\right)+C \] Back-substitute \(\displaystyle z=\sqrt{x}\). Answer: \(\displaystyle \tfrac23x^{3/2}-x+2\sqrt{x}-2\ln\left(\sqrt{x}+1\right)+C\)