Question

$\int \frac{cosx+2sinx}{7sinx-5cosx}dx=ax+bln\mid 7sinx-5cosx\mid +cthena+bis$.

Answer 1

We have,

$\int \frac{\cos x+2\sin x}{7\sin x-5\cos x}dx=ax+b\ln |7\sin x-5\cos x|+C$

Solve it:-

$\int \frac{\cos x+2\sin x}{7\sin x-5\cos x}dx$

We know that,

$\text{Numerator = A}\left(\text{Denominator}\right)+B\left(\text{Derivative}\text{of}\text{Denominator}\right)$……. (1)

Where A and B are two constants.

Now, Differentiation denominator with respect to x and we get

Let,

$7\sin x-5\cos x=t$

$(7\cos x+5\sin x)=\frac{dt}{dx}$

$(7\cos x+5\sin x)dx=dt$

Then,

By equation (1) and we get,

$\cos x+2\sin x=A(7\sin x-5\cos x)+B(7\cos x+5\sin x)$

$=7A\sin x-5A\cos x+7B\cos x+5B\sin x$

$=7A\sin x+5B\sin x-5A\cos x+7B\cos x$

$\cos x+2\sin x=(7A+5B)\sin x+(-5A+7B)\cos x$

Comparing both side and we get,

$7A+5B=1.......\left(2\right)$

$-5A+7B=2.......\left(3\right)$

Solving equation (2) and (3) to, and we get,

$A=-\frac{3}{74},B=\frac{19}{74}$

Put the value of A and B in equation (1) and we get

$\text{Numerator = -}\frac{3}{74}\left(\text{Denominator}\right)+\frac{19}{74}\left(\text{Derivative}\text{of}\text{Denominator}\right)$

$\cos x+2\sin x=-\frac{3}{74}(7\sin x-5\cos x)+\frac{19}{74}(7\cos x+5\sin x)$

Then, According to given question,

$\int \frac{\cos x+2\sin x}{7\sin x-5\cos x}dx=\int \frac{-\frac{3}{74}(7\sin x-5\cos x)+\frac{19}{74}(7\cos x+5\sin x)}{7\sin x-5\cos x}dx$

$=\int \frac{-\frac{3}{74}(7\sin x-5\cos x)}{7\sin x-5\cos x}dx+\frac{\frac{19}{74}(7\cos x+5\sin x)}{7\sin x-5\cos x}dx$

$=-\frac{3}{74}\int \frac{(7\sin x-5\cos x)}{7\sin x-5\cos x}dx+\frac{19}{74}\int \frac{(7\cos x+5\sin x)}{7\sin x-5\cos x}dx$

$=-\frac{3}{74}\int 1dx+\frac{19}{74}\int \frac{dt}{t}$

$\text{On}\text{integrating}\text{and}\text{we}\text{get,}$

$=-\frac{3}{74}x+\frac{19}{74}\ln t$

$=-\frac{3}{74}x+\frac{19}{74}\ln (7\sin x-5\cos x)+C$

$\int \frac{\cos x+2\sin x}{7\sin x-5\cos x}dx=-\frac{3}{74}x+\frac{19}{74}\ln (7\sin x-5\cos x)+C.......\left(4\right)$

Given that,

$\int \frac{\cos x+2\sin x}{7\sin x-5\cos x}dx=ax+b\ln (7\sin x-5\cos x)+C.......\left(5\right)$

On comparing equation (4) and (5) to and we get,

$a=-\frac{3}{74},b=\frac{19}{74}$

Then, $a+b=-\frac{3}{74}+\frac{19}{74}$

$a+b=\frac{16}{74}$

$a+b=\frac{8}{37}$

Hence, this is the answer.