How to solve it ?

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2 Answers
Mar 11, 2018

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

Mar 11, 2018

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.