JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 19)
$$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$ is equal to
(where c is a constant of integration)
(where c is a constant of integration)
2x + sinx + 2sin2x + c
x + 2sinx + sin2x + c
x + 2sinx + 2sin2x + c
2x + sinx + sin2x + c
Explanation
$$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$
= $$\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx $$
= $$\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx$$
= $$\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx$$
Use sin2x = 2sinxcosx and sin3x = 3sinx ā 4sin3x
= $$\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx$$
= $$\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx$$
= $$\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx$$
= $$\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx$$
= x + sin2x + 2sinx + C
= $$\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx $$
= $$\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx$$
= $$\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx$$
Use sin2x = 2sinxcosx and sin3x = 3sinx ā 4sin3x
= $$\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx$$
= $$\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx$$
= $$\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx$$
= $$\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx$$
= x + sin2x + 2sinx + C
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