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

In summary: such as ##\displaystyle \int_0^af(x)dx=\int_0^af(a-x)dx## and some other properties to get a simpler form for the integral.

## 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?

## 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).

haruspex said:
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})##

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?

Change in variable? The variable is still x.​

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.

## 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.

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

Last edited:
SammyS said:
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

## What is a definite integral?

A definite integral is a mathematical concept used in calculus to find the exact area under a curve or between two curves. It is represented by the symbol ∫ and has a lower and upper limit of integration.

## How is a definite integral calculated?

A definite integral is calculated by using the fundamental theorem of calculus, which involves finding the antiderivative of the function being integrated and then evaluating it at the upper and lower limits of integration. The result is the area under the curve between those limits.

## What are the applications of definite integrals?

Definite integrals have many applications in physics, engineering, and economics. They can be used to calculate displacement, velocity, and acceleration in motion problems, as well as to find the total cost, revenue, and profit in business and economics.

## What is the difference between a definite integral and an indefinite integral?

The main difference between a definite integral and an indefinite integral is that a definite integral has specific limits of integration and gives a numerical value, while an indefinite integral has no limits and gives a general algebraic expression.

## Can definite integrals be used to solve real-life problems?

Yes, definite integrals are commonly used to solve real-life problems in various fields such as engineering, physics, and economics. They can be used to calculate the area under a curve, which can represent quantities such as displacement, velocity, or profit. This allows for more accurate and precise calculations in real-world scenarios.

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