-cos4 x dx-sin3 xsin5 x - cos5 x3-5 - - 1-2 1 - cotA xB - C then AB-

I = ∫cosx^4dx/sin^3 x(sin^5 x + cos^5 x)^3/5 = ∫cosec^2 xcot^4 xdx/(1+cot^5 x)^3/5 1 + cot^5 x = t. 5 cot^4 x ( cosec^2 x)dx = dt = 1/5∫dt/t^3/5 = t ...

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Byju's Answer

Standard XIII

Mathematics

Integration of Trigonometric Functions

∫cos4 x dx/si...

Question

$\int \frac{c o s^{4} x d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = - \frac{1}{2} (1 + c o t^{A} x)^{B} + C$ then AB=

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\frac{1}{2}

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None of these

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Solution

The correct option is B 2
$I = \int \frac{c o s x^{4} d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = \int \frac{c o s e c^{2} x c o t^{4} x d x}{(1 + c o t^{5} x)^{3 / 5}} 1 + c o t^{5} x = t . 5 c o t^{4} x (- c o s e c^{2} x) d x = d t = \frac{- 1}{5} \int \frac{d t}{t^{3 / 5}} = \frac{- t^{2 / 5}}{2} = \frac{- 1}{2} (1 + c o t^{5} x)^{2 / 5} + C .$

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Similar questions

\int \frac{c o s^{4} x d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = - \frac{1}{2} (1 + c o t^{A} x)^{B} + C

then AB=

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∫cos4xdxsin3x(sin5x+cos5x)35=−12(1+cotAx)B+C then AB=

The correct option is B 2I=∫cosx4dxsin3x(sin5x+cos5x)35=∫cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5 cot4x(−cosec2x)dx=dt=−15∫dtt3/5=−t2/52=−12(1+cot5x)2/5+C.

$\int \frac{c o s^{4} x d x}{s i n^{3} x (s i n^{5} x + c o s^{5} x)^{\frac{3}{5}}} = - \frac{1}{2} (1 + c o t^{A} x)^{B} + C$ then AB=