Find the integrals of the functions- -cos 2 x cos 4 x cos 6 x d x

∫ cos 2x cos 4x cos 6x dx =∫ cos 2x [ 1/2{ cos (4x+6x)+cos (4x 6x) } ]dx [∴ 2 cos A cos B =cos (A+B)+cos (A B)] =1/2∫ [cos 2x cos 10x +cos 2x cos ( 2x)]d ...

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Byju's Answer

Standard XII

Mathematics

Argument of a Complex Number

Find the inte...

Question

Find the integrals of the functions.
$\int c o s 2 x$ cos 4x cos 6x dx.

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Solution

$\int c o s 2 x c o s 4 x c o s 6 x d x = \int c o s 2 x [\frac{1}{2} {c o s (4 x + 6 x) + c o s (4 x - 6 x)}] d x [∴ 2 c o s A c o s B = c o s (A + B) + c o s (A - B)] = \frac{1}{2} \int [c o s 2 x c o s 10 x + c o s 2 x c o s (- 2 x)] d x = \frac{1}{2} \int [c o s 2 x c o s 10 x + c o s^{2} 2 x] d x [∵ c o s (- θ) = c o s θ] = \frac{1}{2} \int [\frac{1}{2} (c o s (2 x + 10 x) + c o s (2 x - 10 x)) + (\frac{1 + c o s 4 x}{2})] d x [∴ c o s 2 θ = 2 c o s^{2} θ - 1] = \frac{1}{4} \int [c o s 12 x + c o s 8 x + 1 + c o s 4 x] d x = \frac{1}{4} [\frac{s i n 12 x}{12} + \frac{s i n 8 x}{8} + x + \frac{s i n 4 x}{4}] + C$

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Find the integrals of the functions. ∫cos2x cos 4x cos 6x dx.

Find the integrals of the functions.
$\int c o s 2 x$ cos 4x cos 6x dx.