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NCERT Solutions · Class 12 Mathematics Integrals

261 questions · 261 still being checked

EXERCISE 7.3 21–24 (part 9 of 27)

  1. Find the integrals of the functions in Exercises $\displaystyle 1$ to $\displaystyle 22$:

    Exercise 21

    sin⁡−1(cos⁡x)\displaystyle \sin ^{-1}(\cos x)

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    NCERT’s answer
    \(\displaystyle \frac{\pi x}{2}-\frac{x^{2}}{2}+\mathrm{C}\)
    First simplify \(\displaystyle \sin^{-1}(\cos x)\); this is where the domain matters. Write \(\displaystyle \cos x=\sin\left(\dfrac{\pi}{2}-x\right)\). The identity \(\displaystyle \sin^{-1}(\sin\theta)=\theta\) is valid only when \(\displaystyle \theta\) lies in the principal range \(\displaystyle \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]\), and \(\displaystyle \dfrac{\pi}{2}-x\) lies there exactly when \(\displaystyle 0\le x\le\pi\). So, for \(\displaystyle x\in[0,\pi]\), \[\sin^{-1}(\cos x)=\sin^{-1}\!\left(\sin\left(\frac{\pi}{2}-x\right)\right)=\frac{\pi}{2}-x.\] Integrating this linear function, \[\int\sin^{-1}(\cos x)\,dx=\int\left(\frac{\pi}{2}-x\right)dx=\frac{\pi x}{2}-\frac{x^{2}}{2}+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{\pi x}{2}-\dfrac{x^{2}}{2}+\mathrm{C}\), on \(\displaystyle 0\le x\le\pi\) (outside that interval the simplification, and hence the formula, changes).

  2. Exercise 22

    1cos⁡(x−a)cos⁡(x−b)\displaystyle \frac{1}{\cos (x-a) \cos (x-b)}

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    NCERT’s answer
    \(\displaystyle \frac{1}{\sin (a-b)} \log \left|\frac{\cos (x-a)}{\cos (x-b)}\right|+\mathrm{C}\)
    The two angles differ by a constant: \(\displaystyle (x-b)-(x-a)=a-b\). Multiply and divide by \(\displaystyle \sin(a-b)\) (a non-zero constant when \(\displaystyle a\neq b\)) and expand it as the sine of that difference: \[\sin(a-b)=\sin\big[(x-b)-(x-a)\big]=\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a).\] Dividing this equation by \(\displaystyle \cos(x-a)\cos(x-b)\) gives \[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}=\tan(x-b)-\tan(x-a),\] so \[\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\big[\tan(x-b)-\tan(x-a)\big].\] Now integrate, using \(\displaystyle \int\tan(x-k)\,dx=-\log|\cos(x-k)|\): \[\int\frac{dx}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\Big[-\log|\cos(x-b)|+\log|\cos(x-a)|\Big]+\mathrm{C}.\] Final answer: \(\displaystyle \dfrac{1}{\sin(a-b)}\log\left|\dfrac{\cos(x-a)}{\cos(x-b)}\right|+\mathrm{C}\).
  3. Choose the correct answer in Exercises $\displaystyle 23$ and 24.

    Exercise 23

    ∫sin⁡2x−cos⁡2xsin⁡2xcos⁡2xdx\displaystyle \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x is equal to (A) tan⁡x+cot⁡x+C\displaystyle \tan x+\cot x+\mathrm{C} (B) tan⁡x+cosec⁡x+C\displaystyle \tan x+\operatorname{cosec} x+\mathrm{C} (C) −tan⁡x+cot⁡x+C\displaystyle -\tan x+\cot x+\mathrm{C} (D) tan⁡x+sec⁡x+C\displaystyle \tan x+\sec x+\mathrm{C}

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    NCERT’s answer
    A
    Split the single fraction into two and cancel in each: \[\frac{\sin^{2}x-\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}-\frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{1}{\cos^{2}x}-\frac{1}{\sin^{2}x}=\sec^{2}x-\mathrm{cosec}^{2}x.\] Use \(\displaystyle \int\sec^{2}x\,dx=\tan x\) and \(\displaystyle \int\mathrm{cosec}^{2}x\,dx=-\cot x\); the second minus sign meets that one and gives a plus: \[\int\left(\sec^{2}x-\mathrm{cosec}^{2}x\right)dx=\tan x-(-\cot x)+\mathrm{C}=\tan x+\cot x+\mathrm{C}.\] Final answer: option (A), \(\displaystyle \tan x+\cot x+\mathrm{C}\).

  4. Exercise 24

    ∫ex(1+x)cos⁡2(exx)dx\displaystyle \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x equals (A) −cot⁡(exx)+C\displaystyle -\cot \left(e x^{x}\right)+\mathrm{C} (B) tan⁡(xex)+C\displaystyle \tan \left(x e^{x}\right)+\mathrm{C} (C) tan⁡(ex)+C\displaystyle \tan \left(e^{x}\right)+\mathrm{C} (D) cot⁡(ex)+C\displaystyle \cot \left(e^{x}\right)+\mathrm{C}

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    NCERT’s answer
    B
    Read the numerator as a derivative. By the product rule, \[\frac{d}{dx}\left(x e^{x}\right)=e^{x}+x e^{x}=e^{x}(1+x),\] which is exactly the numerator, and the angle in the denominator is that same \(\displaystyle x e^{x}\). Substitute \(\displaystyle t=x e^{x}\), so \(\displaystyle dt=e^{x}(1+x)\,dx\): \[\int\frac{e^{x}(1+x)}{\cos^{2}\left(x e^{x}\right)}\,dx=\int\frac{dt}{\cos^{2}t}=\int\sec^{2}t\,dt=\tan t+\mathrm{C}.\] Putting \(\displaystyle t=x e^{x}\) back: Final answer: option (B), \(\displaystyle \tan\left(x e^{x}\right)+\mathrm{C}\).