Verified Solution • 2025–26 Session moderate

QJEEM19Ma8c80: JEE Main 2019 Mathematics - Class 12 JEEM19Ma8c80 Solution

∫sin⁡5x2sin⁡x2dx\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} is equal to
(where c is a constant of integration)

Multiple Choice Options

  • A 2x + sinx + 2sin2x + c
  • B x + 2sinx + sin2x + c
  • C x + 2sinx + 2sin2x + c
  • D 2x + sinx + sin2x + c

Step-by-Step Verified Solution

∫sin⁡5x2sin⁡x2dx\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx}

= ∫2cos⁡x2.sin⁡5x22cos⁡x2.sin⁡x2dx\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx

= ∫sin⁡(5x2+x2)+sin⁡(5x2−x2)sin⁡xdx\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx

= ∫sin⁡3x+sin⁡2xsin⁡xdx\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx

Use sin2x = 2sinxcosx and sin3x = 3sinx – 4sin3x

= ∫3sin⁡x−4sin⁡3x+2sin⁡xcos⁡xsin⁡xdx\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx

= ∫(3−4sin⁡2x+2cos⁡x)dx\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx

= ∫(3−2(1−cos⁡2x)+2cos⁡x)dx\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx

= ∫(1+2cos⁡2x+2cos⁡x)dx\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx

= x + sin2x + 2sinx + C

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring