-0-4 sin x-cos x-9-16 sin 2xdx-

Let I=∫_0^π/4 sin x+cos x/9+16 sin 2xdx Put sin x cos x=t, then (sin x+cos x)dx=dt I=∫_ 1^0 dt/9+16(1 t^2)=∫_ 1^0 dt/25 16t^2 =1/10∫_ 1^0 ( 1/5 4t+1/5+4t )dt = ...

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Byju's Answer

Standard XI

Mathematics

Property 1

∫0π/4 sin x+c...

Question

$\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x =$

\frac{1}{20} l o g 3

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l o g 3

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\frac{1}{20} l o g 5

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{1}{20} l o g 3$
$L e t I = \int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x$
$P u t s i n x - c o s x = t, t h e n (s i n x + c o s x) d x = d t$
$I = \int_{- 1}^{0} \frac{d t}{9 + 16 (1 - t^{2})} = \int_{- 1}^{0} \frac{d t}{25 - 16 t^{2}}$
$= \frac{1}{10} \int_{- 1}^{0} (\frac{1}{5 - 4 t} + \frac{1}{5 + 4 t}) d t$
$= {∣ ∣ \frac{1}{10} . \frac{1}{4} [l o g (5 + 4 t) - l o g (5 - 4 t)] ∣ ∣}_{- 1}^{0}$
$= \frac{1}{40} (l o g 9 - l o g 1) = \frac{1}{20} l o g 3$

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\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x =

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\int_{0}^{π / 4} \frac{sin x + cos x}{9 + 16 sin 2 x} d x

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∫π40 sin x+cos x9+16 sin 2xdx=

The correct option is A 120log 3Let I=∫π40 sin x+cos x9+16 sin 2xdx Put sin x−cos x=t, then (sin x+cos x)dx=dt I=∫0−1 dt9+16(1−t2)=∫0−1dt25−16t2 =110∫0−1(15−4t+15+4t)dt =∣∣110.14[log(5+4t)−log(5−4t)]∣∣0−1 =140(log 9−log 1)=120log 3

$\int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{9 + 16 s i n 2 x} d x =$