Question

How to solve it ?

Answer 1

`I=1/2(x+ln(cos(x)+sin(x)))+C`

Explanation:

We want to solve

`I=intcos(x)/(cos(x)+sin(x))dx`

We can quite easily find

`color(green)(I_1=int(cos(x)+sin(x))/(cos(x)+sin(x))dx` and `color(green)(I_2=int(cos(x)-sin(x))/(cos(x)+sin(x))dx)`

So can we find some constants such

`I=AI_1+BI_2`

Only the denominator have changed, thus we seek

`cos(x)=A(cos(x)+sin(x))+B(cos(x)-sin(x))`

By letting `x=0` and `x=pi/2`

`1=A+B`
`color(white)(0=A-Bsscdcfcss)=>A=B=1/2`
`0=A-B`

Therefore

`I=1/2int(cos(x)+sin(x))/(cos(x)+sin(x))dx+1/2int(cos(x)-sin(x))/(cos(x)+sin(x))dx`

`color(white)(I)=1/2intdx+1/2int(1)/(u)du`

`color(white)(I)=1/2x+1/2ln(u)+C`

`color(white)(I)=1/2(x+ln(cos(x)+sin(x)))+C`

The other can be solve by similar approach

Answer 2

`I_1=1/2[x+ln|(sinx+cosx)|]+C_1,`

and

`I_2=1/2[x-ln|(sinx+cosx)|]+C_2`.

Explanation:

`I_1=intcosx/(cosx+sinx)dx, and I_2=intsinx/(sinx+cosx)dx`.

`:. I_1+I_2=int(cosx+sinx)/(sinx+cosx)dx=int1dx`.

`:. I_1+I_2=x..............................................(ast^1)`.

Also, `I_1-I_2=int(cosx-sinx)/(sinx+cosx)dx`,

`=int{d/dx(sinx+cosx)}/(sinx+cosx)dx`.

`:. I_1-I_2=ln|(sinx+cosx)|........................(ast^2)`.

Soving `I_1 and I_2`, we have,

`I_1=1/2[x+ln|(sinx+cosx)|]+C_1, and,`

`I_2=1/2[x-ln|(sinx+cosx)|]+C_2`.