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Definite Integration question

2019 · 10 Apr · Shift 2 · Q37
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Definite Integration question

2019 · 10 Apr · Shift 2 · Q37

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The integral ∫π/6π/3sec⁡2/3xcos⁡ec4/3xdx\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}} x\cos e{c^{4/3}}xdxπ/6∫π/3​sec2/3xcosec4/3xdx is equal to :
  1. A
    353−313{3^{{5 \over 3}}} - {3^{{1 \over 3}}}335​−331​
  2. B
    356−323{3^{{5 \over 6}}} - {3^{{2 \over 3}}}365​−332​
  3. C
    343−313{3^{{4 \over 3}}} - {3^{{1 \over 3}}}334​−331​
  4. D
    376−356{3^{{7 \over 6}}} - {3^{{5 \over 6}}}367​−365​
View written solutionFree

Correct answer: D

  1. Interpret the integrand carefully

    The given integral is ∫π/6π/3sec⁡2/3x cosec⁡4/3x dx.\int_{\pi/6}^{\pi/3} \sec^{2/3}x\,\cosec^{4/3}x\,dx.∫π/6π/3​sec2/3xcosec4/3xdx.

    So we need to evaluate I=∫π/6π/3sec⁡2/3x cosec⁡4/3x dx.I=\int_{\pi/6}^{\pi/3} \sec^{2/3}x\,\cosec^{4/3}x\,dx.I=∫π/6π/3​sec2/3xcosec4/3xdx.

  2. Rewrite in terms of sine and cosine

    Using sec⁡x=1cos⁡x,cosec⁡x=1sin⁡x,\sec x=\frac{1}{\cos x},\qquad \cosec x=\frac{1}{\sin x},secx=cosx1​,cosecx=sinx1​, we get

    =\frac{1}{\cos^{2/3}x\sin^{4/3}x}.$$ Now factor this as $$\frac{1}{\cos^{2/3}x\sin^{4/3}x} =\frac{\sin^{2/3}x}{\cos^{2/3}x\sin^2 x} =\tan^{2/3}x\,\cosec^2 x.$$ Hence $$I=\int_{\pi/6}^{\pi/3} \tan^{2/3}x\,\cosec^2x\,dx.$$

  3. Substitute

    Let t=cot⁡x.t=\cot x.t=cotx. Then dt=−cosec⁡2x dx,dt=-\cosec^2x\,dx,dt=−cosec2xdx, and since tan⁡x=1cot⁡x=1t,\tan x=\frac{1}{\cot x}=\frac{1}{t},tanx=cotx1​=t1​, we have tan⁡2/3x=t−2/3.\tan^{2/3}x=t^{-2/3}.tan2/3x=t−2/3.

    Therefore,

    =-\int t^{-2/3}\,dt.$$

  4. Change the limits

    When x=π/6x=\pi/6x=π/6, t=cot⁡π6=3.t=\cot\frac{\pi}{6}=\sqrt{3}.t=cot6π​=3​.

    When x=π/3x=\pi/3x=π/3, t=cot⁡π3=13.t=\cot\frac{\pi}{3}=\frac{1}{\sqrt{3}}.t=cot3π​=3​1​.

    So

    =\int_{1/\sqrt{3}}^{\sqrt{3}} t^{-2/3}\,dt.$$

  5. Integrate

    ∫t−2/3 dt=t1/31/3=3t1/3.\int t^{-2/3}\,dt=\frac{t^{1/3}}{1/3}=3t^{1/3}.∫t−2/3dt=1/3t1/3​=3t1/3.

    Thus

    =3\left((\sqrt{3})^{1/3}-\left(\frac{1}{\sqrt{3}}\right)^{1/3}\right).$$

  6. Simplify powers

    3=31/2  ⟹  (3)1/3=31/6,\sqrt{3}=3^{1/2} \implies (\sqrt{3})^{1/3}=3^{1/6},3​=31/2⟹(3​)1/3=31/6, and (13)1/3=3−1/6.\left(\frac{1}{\sqrt{3}}\right)^{1/3}=3^{-1/6}.(3​1​)1/3=3−1/6.

    Hence

    =3^{7/6}-3^{5/6}.$$

  7. Match with options

    This is exactly Option D: 37/6−35/6.\boxed{3^{7/6}-3^{5/6}}.37/6−35/6​.

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