Question

evaluate :[2sinx+3cosx/7sinx-2cosx]dx​

Answer 1

Step-by-step explanation:

We have I=∫

3cosx+2sinx

3sinx+2cosx

dx

Let 3sinx+2cosx=λ(3cosx+2sinx)+μ

dx

d

(3cosx+2sinx)

3sinx+2cosx=λ(3cosx+2sinx)+μ(−3sinx+2cosx)

Comparing the coefficients of sinx and cosx on both sides,we get

−3μ+2λ=3and2μ+3λ=2

λ=

13

12

andμ=−

13

5

therefore I=∫

3cosx+2sinx

λ(3cosx+2sinx)+μ(−3sinx+2cosx)

dx

=λ∫1dx+μ∫

3cosx+2sinx

−3sinx+2cosx

dx

Put t=3cosx+2sinx

⇒dt=−3sinx+2cosx

=λx+μ∫

t

dt

=λx+μlog∣t∣+C

=

13

12

x+

13

−5

log∣3cosx+2sinx∣+C