How to Simplify the Integral of (sinx-cosx)ln(sinx) from 0 to π/2?

[AdityaDev]

Homework Statement

$$\int_0^{\pi/2}(sinx-cosx)ln(sinx)dx$$

Homework Equations

$int_0^af(x)dx=int_0^af(a-x)dx$

The Attempt at a Solution

Using above equation, you get (without integral sign):
$(sinx-cosx)ln(tanx)$ but it did not make any difference.
I got the answer by splitting the integral to $sinxln(sinx)$ and $cosxln(cosx)$ and then using integration by parts.(Answer verified without limits in wolfram alpha)
But the answer was long and time taking. Also the question is objective type and there has to be a shorter method. Can you provide a shorter method?

 

[haruspex]

Homework Statement

$$\int_0^{\pi/2}(sinx-cosx)ln(sinx)dx$$

Homework Equations

$int_0^af(x)dx=int_0^af(a-x)dx$

The Attempt at a Solution

Using above equation, you get (without integral sign):
$(sinx-cosx)ln(tanx)$ but it did not make any difference.
I got the answer by splitting the integral to $sinxln(sinx)$ and $cosxln(cosx)$ and then using integration by parts.(Answer verified without limits in wolfram alpha)
But the answer was long and time taking. Also the question is objective type and there has to be a shorter method. Can you provide a shorter method?

There's a way to express sin(x)-cos(x) as A sin(y).

 

[AdityaDev]

There's a way to express sin(x)-cos(x) as A sin(y).

Yes. Multiply and divide by $\sqrt{2}$ and put $\frac{1}{\sqrt{2}}$ as cos or sine to get $cosAsinB-sinAcosB$ which can be expressed as $sin(B-A)$.
So $sinx-cosx=\sqrt{2}sin(x-\frac{\pi}{4})$

 

[haruspex]

Yes. Multiply and divide by $\sqrt{2}$ and put $\frac{1}{\sqrt{2}}$ as cos or sine to get $cosAsinB-sinAcosB$ which can be expressed as $sin(B-A)$.
So $sinx-cosx=\sqrt{2}sin(x-\frac{\pi}{4})$

Right. Does that suggest a change of variable?

 

[AdityaDev]

Change in variable? The variable is still x.​

 

[haruspex]

Change in variable? The variable is still x.​

I was suggesting you make a change of variable to simplify the sin expression, but what I had in mind doesn't work.
Another way is to go to half angles, x=2u. In the original integral, the cos(x)ln(sin(x)).dx term is no problem, since that's like ln(y)dy. Expanding sin(2u) in the other term gets you to integrals like y.ln(y)dy. Whether this is simpler than the method you found is doubtful.

 

[SammyS]

Homework Statement

$$\int_0^{\pi/2}(sinx-cosx)ln(sinx)dx$$

Homework Equations

$\displaystyle \int_0^af(x)dx=\int_0^af(a-x)dx$

...

I assume you are to evaluate the given definite integral by making use of the integral equivalence that's given.

You then have a = π/2 , so that $\displaystyle f(a - x) = \left(\sin(\pi/2-x)-\cos(\pi/2-x)\right)\ln\left(\sin(\pi/2-x)\right)$

However, $\sin(\pi/2-x)=\cos(x)\$ and $\cos(\pi/2-x)=\sin(x)\$ .

Make those substitutions & see what you get.

Added in Edit:

It looks like this is what you did to get $(\sin x-\cos x)\ln(\tan x)\$, so, never mind.

 

[AdityaDev]

I assume you are to evaluate the given definite integral by making use of the integral equivalence that's given.

You then have a = π/2 , so that $\displaystyle f(a - x) = \left(\sin(\pi/2-x)-\cos(\pi/2-x)\right)\ln\left(\sin(\pi/2-x)\right)$

However, $\sin(\pi/2-x)=\cos(x)\$ and $\cos(\pi/2-x)=\sin(x)\$ .

it does not make much difference. I have to make use properties of definite integrals