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Indefinite Integrals question

2019 · 9 Apr · Shift 1 · Q40
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Indefinite Integrals question

2019 · 9 Apr · Shift 1 · Q40

JEE MainMathematicsIndefinite IntegralsMCQ+4 / −1
The integral ∫sec2/3 x cosec4/3x dx\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}}∫sec2/3xcosec4/3xdx is equal to (Hence C is a constant of integration)
  1. A
    -3/4 tan - 4 / 3 x + C
  2. B
    3tan–1/3x + C
  3. C
    –3cot–1/3x+ C
  4. D
    - 3tan–1/3x + C
View written solutionFree

Correct answer: D

  1. Rewrite the integrand in terms of sine and cosine

Given I=∫sec⁡2/3x csc⁡4/3x dxI=\int \sec^{2/3}x\,\csc^{4/3}x\,dxI=∫sec2/3xcsc4/3xdx

Using sec⁡x=1cos⁡x,csc⁡x=1sin⁡x\sec x=\frac{1}{\cos x},\qquad \csc x=\frac{1}{\sin x}secx=cosx1​,cscx=sinx1​ we get I=∫dxcos⁡2/3x sin⁡4/3x.I=\int \frac{dx}{\cos^{2/3}x\,\sin^{4/3}x}.I=∫cos2/3xsin4/3xdx​.

Now factor this as

=\frac{1}{\sin^{2}x}\cdot \frac{\sin^{2/3}x}{\cos^{2/3}x} =\csc^2 x\,\tan^{2/3}x.$$ So, $$I=\int \tan^{2/3}x\,\csc^2x\,dx.$$ --- 2. **Substitute using cotangent** Let $$u=\cot x.$$ Then $$du=-\csc^2x\,dx \quad\Rightarrow\quad \csc^2x\,dx=-du.$$ Also, $$\tan x=\frac{1}{\cot x}=\frac{1}{u}$$ so $$\tan^{2/3}x=u^{-2/3}.$$ Therefore, $$I=\int u^{-2/3}(-du)=-\int u^{-2/3}du.$$ --- 3. **Integrate** $$-\int u^{-2/3}du =-\left(\frac{u^{1/3}}{1/3}\right)+C =-3u^{1/3}+C.$$ Substituting back $u=\cot x$, $$I=-3\cot^{1/3}x+C.$$ --- 4. **Match with options** This is exactly $$-3\cot^{-1/3}x+C$$ if the notation in the options means $\dfrac{-3}{\cot^{1/3}x}$? But from standard reading, option C is intended as $$-3\cot^{-1/3}x+C=-\frac{3}{\cot^{1/3}x}+C=-3\tan^{1/3}x+C,$$ which is **not** what we obtained. So the correct antiderivative is $$\boxed{-3\cot^{1/3}x+C}.$$ Since none of the options is written cleanly, let us check option D: $$-3\tan^{-1/3}x+C=-3\left(\tan x\right)^{-1/3}+C=-3\cot^{1/3}x+C,$$ which matches our result. Hence the correct option is **D**. --- 5. **Verification by differentiation** Differentiate $$F(x)=-3\tan^{-1/3}x.$$ Then $$F'(x)=-3\left(-\frac13\right)\tan^{-4/3}x\sec^2x =\tan^{-4/3}x\sec^2x.$$ Now, $$\tan^{-4/3}x\sec^2x =\frac{\sec^2x}{\tan^{4/3}x} =\frac{1/\cos^2x}{(\sin x/\cos x)^{4/3}} =\frac{1}{\cos^{2/3}x\sin^{4/3}x} =\sec^{2/3}x\csc^{4/3}x.$$ So the result is confirmed.
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