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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

∫ ^ 8 x - ^ 8 x 1 - 2 ^ 2 x ^ 2 x dx is equal to (where C is an...

∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2xdx∫ \dfrac{\sin^{8} \text{x} - \cos^{8} \text{x}}{1 - 2\sin^{2} \text{x}\cos^{2} \text{x}} \text{dx} is equal to (where CC is an arbitrary constant)

Options

  1. A12sin⁡2x+C\dfrac{1}{2} \sin 2x + \text{C}
  2. B−12sin⁡2x+C- \dfrac{1}{2} \sin 2x + \text{C}
  3. C−12sin⁡x+C- \dfrac{1}{2} \sin x + \text{C}
  4. D−sin⁡2x+C- \sin^{2} \text{x} + \text{C}

Correct answer

B. −12sin⁡2x+C- \dfrac{1}{2} \sin 2x + \text{C}

Step-by-step solution

∫sin⁡8x−cos⁡8x1−2sin2xcos2xdx∫ \dfrac{\sin^{8}\text{x} - \cos^{8}\text{x}}{1 - {2sin}^{2}{xcos}^{2}\text{x}}\text{dx}

=∫(sin⁡4x−cos⁡4x)(sin⁡4x+cos4x)1−2sin2xcos2xdx= ∫ \dfrac{\left(\sin^{4}\text{x} - \cos^{4}\text{x}\right)\left(\sin^{4}{x + cos}^{4}\text{x}\right)}{1 - {2sin}^{2}{xcos}^{2}\text{x}}\text{dx}

=∫(sin⁡2x−cos⁡2x)(sin⁡4x+cos4x)1−2sin2xcos2xdx= ∫ \dfrac{\left(\sin^{2}\text{x} - \cos^{2}\text{x}\right)\left(\sin^{4}{x + cos}^{4}\text{x}\right)}{1 - {2 sin}^{2}{xcos}^{2}\text{x}}\text{dx}

= ∫ - \text{\cos 2x dx} = - \dfrac{1}{2}\text{\sin 2x + C}

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