Consider the integral I -010 -cos 6 x -cos 7 x -cos 8 x -cos 9 x-1- e 2 sin 3 4 x d xIf I-K -0-2cos 6 x -cos 7 x -cos 8 x -cos 9 x d x then -k-is equal toA- 1B- 10C- 20D- 5

The correct option is B 10I=∫_0^10πcos6x.cos7x.cos8x.cos9x/1+e^2sin^34xdx…(1) I=∫_0^10πcos6x.cos7x.cos8x.cos9x/1+e^ 2sin^34xdx (using ∫_0^af(x)=∫_0^af(a x)) ...

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Byju's Answer

Standard XII

Mathematics

Property 6

Consider the ...

Question

Consider the integral $I = \int_{0}^{10 π} \frac{c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x}{1 + e^{2 s i n^{3} 4 x}} d x$

If $I = K \int_{0}^{\frac{π}{2}} c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x d x$ then 'k'is equal to

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Solution

The correct option is C 10
$I = \int_{0}^{10 π} \frac{c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x}{1 + e^{2 s i n^{3} 4 x}} d x \dots (1) I = \int_{0}^{10 π} \frac{c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x}{1 + e^{- 2 s i n^{3} 4 x}} d x (using \int_{0}^{a} f (x) = \int_{0}^{a} f (a - x)) I = \int_{0}^{10 π} \frac{e^{2 s i n^{3} 4 x} (c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x)}{1 + e^{2 s i n^{3} 4 x}} d x \dots (2) Add (1) + (2) 2 I = \int_{0}^{10 π} c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x d x$
Let $f (x) = c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x$

$f (x) repeats after π interval 2 I = 5 \int_{0}^{2 π} f (s) d x; a g a i n f (2 a - x) = f (x) 2 I = 10 \int_{0}^{π} f (x) d x; a g a i n f (π - x) = f (x) 2 I = 20 \int_{0}^{\frac{π}{2}} f (x) d x \Rightarrow I = 10 \int_{0}^{\frac{π}{2}} f (x) d x ∴ I = 10 \int_{0}^{\frac{π}{2}} c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x d x$

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Consider the integral I=∫10π0cos6x.cos7x.cos8x.cos9x1+e2sin34xdx If I=K∫π20cos6x.cos7x.cos8x.cos9xdx then 'k'is equal to

Consider the integral $I = \int_{0}^{10 π} \frac{c o s 6 x . c o s 7 x . c o s 8 x . c o s 9 x}{1 + e^{2 s i n^{3} 4 x}} d x$

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