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Question

sin3x.cos4x dx


A

cos7(x)7cos5(x)5+c

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B

cos7(x)5+cos5(x)7+c

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C

cos7(x)3+cos5(x)5+c

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D

cos7(x)5+cos5(x)3+c

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Solution

The correct option is A

cos7(x)7cos5(x)5+c


We can see here that the above expression is in sinmx.cosnx dx form with having one power odd and another even. We saw that in such case we substitute the one which has even power.

Since cos(x) has even power we’ll substitute cos (x) = t .

sin3x.cos4x dx

Orsin2(x).sin(x).cos4(x)dx

Let’s substitute cos(x) = t

- sin(x). dx= dt

So, we’ll have -

(1t2).t4 dt

Or t4dt+ t6dt

t77t55+c

Let’s substitute t = cos(x)

And we’ll have

cos7(x)7cos5(x)5+c


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