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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

∫ ^ 4 x ^ 8 x dx is equal to (where C is an arbitrary constant)

∫sin⁡4xcos⁡8xdx∫ \dfrac{\sin^{4} x}{\cos^{8} x}dx is equal to (where CC is an arbitrary constant)

Options

  1. A(1+tan⁡5x)5+tan⁡5x7+C\dfrac{\left(1 + \tan^{5} x\right)}{5} + \dfrac{\tan^{5} x}{7} + C
  2. Btan⁡5x5+tan⁡7x7+C\dfrac{\tan^{5} x}{5} + \dfrac{\tan^{7} x}{7} + C
  3. Ctan⁡7x5+tan⁡5x7+C\dfrac{\tan^{7} x}{5} + \dfrac{\tan^{5} x}{7} + C
  4. DNone of these

Correct answer

B. tan⁡5x5+tan⁡7x7+C\dfrac{\tan^{5} x}{5} + \dfrac{\tan^{7} x}{7} + C

Step-by-step solution

∫sin⁡4xcos⁡4xcos⁡8xcos⁡4x.dx=∫tan⁡4x.sec⁡4xdx∫ \dfrac{\dfrac{\sin^{4} x}{\cos^{4} x}}{\dfrac{\cos^{8} x}{\cos^{4} x}}. dx = ∫ \tan^{4} x. \sec^{4} xdx

=∫tan⁡4x(1+tan⁡2x)sec⁡2xdx= ∫ \tan^{4} x\left(1 + \tan^{2} x\right)\sec^{2} xdx

=∫tan⁡4x(1+tan⁡2x)d(tan⁡x)=∫t4(1+t2)dt=t55+t77+C= ∫ \tan^{4} x\left(1 + \tan^{2} x\right)d\left(\tan x\right) = ∫ t^{4}{\left(1 + t^{2}\right)}^{} dt= \dfrac{t^{5}}{5} + \dfrac{t^{7}}{7} + \text{C}
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