Question

$∫9cosx - sinx \\ 4sinx + 5cosx dx$
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Answer 1

Step-by-step explanation:

Consider the given integral.

I=∫

4sinx+5cosx

9cosx−sinx

dx

I=∫

4sinx+5cosx

4sinx+5cosx−5sinx+4cosx

dx

I=∫(1)dx+∫

4sinx+5cosx

4cosx−5sinx

dx

I=I

1

+I

2

Therefore,

I

2

=∫

4sinx+5cosx

4cosx−5sinx

dx

Let t=4sinx+5cosx

4sinx+5cosx

dt=(4cosx−5sinx)dx

Therefore,

I

2

=∫

t

1

dt

I

2

=log(4sinx+5cosx)+C

So,

I=x+log(4sinx+5cosx)+C

Hence. this is the answer.