-0-2cos x-1-sin x2-sin x d x-U P S E A T 1999-

The correct option is A Put sin x = t ⇒ cos x dx = dt, so reduced integral is ∫_0^1 ( 1/1+t 1/2+t )dt = [log(1+t) log(2+t)]^1_0 = log 2/3 log 1/2 = log 4/3 ...

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Standard XII

Mathematics

Property 1

∫0π/2cos x/1+...

Question

$\int_{0}^{\frac{π}{2}} \frac{c o s x}{(1 + s i n x) (2 + s i n x)} d x =$ [UPSEAT 1999]

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Solution

The correct option is A
Put sin x = t $\Rightarrow$ cos x dx = dt, so reduced integral is $\int_{0}^{1} (\frac{1}{1 + t} - \frac{1}{2 + t}) d t = [l o g (1 + t) - l o g (2 + t)]_{0}^{1} = l o g \frac{2}{3} - l o g \frac{1}{2} = l o g \frac{4}{3}$

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Similar questions

Q. Evaluate:

\int_{0}^{π / 2} \frac{cos x}{(1 + sin x) (2 + sin x)} d x

\int_{0}^{π / 2} \frac{cos x}{(1 + sin x) (2 + sin x)} d x

is equal to

\int_{0}^{π_{/} 2} \frac{cos x}{1 + sin x} d_{X =}

π / 2 \int 0 \frac{cos x}{1 + cos x + sin x} d x

\int_{0}^{\frac{π}{2}} \frac{c o s x}{(1 + s i n x) (2 + s i n x)} d x =

[UPSEAT 1999]

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∫π20 cos x(1+sin x)(2+sin x)dx= [UPSEAT 1999]

The correct option is A Put sin x = t ⇒ cos x dx = dt, so reduced integral is ∫10(11+t−12+t)dt=[log(1+t)−log(2+t)]10=log23−log12=log43

$\int_{0}^{\frac{π}{2}} \frac{c o s x}{(1 + s i n x) (2 + s i n x)} d x =$ [UPSEAT 1999]

The correct option is A
Put sin x = t $\Rightarrow$ cos x dx = dt, so reduced integral is $\int_{0}^{1} (\frac{1}{1 + t} - \frac{1}{2 + t}) d t = [l o g (1 + t) - l o g (2 + t)]_{0}^{1} = l o g \frac{2}{3} - l o g \frac{1}{2} = l o g \frac{4}{3}$