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Question: What is the integration of \(\int{\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)}dx}\) ?...

What is the integration of ∫cos⁡x+xsin⁡xx(x+cos⁡x)dx\int{\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)}dx} ?

Explanation

Solution

Hint: Try to break the numerator into (x+cos⁡x)−x(1−sin⁡x)\left( x+\cos x \right)-x\left( 1-\sin x \right) then break it. Apply integration separately and you’ll get the answer.

Complete step by step answer:

In the fraction cos⁡x+xsin⁡xx(x+cos⁡x)\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)} we will consider numerator first which is cos⁡x+xsin⁡x\cos x+x\sin x which can be written as (x+cos⁡x−x+xsin⁡x)\left( x+\cos x-x+x\sin x \right) which can be further written as
(x+cos⁡x−(x−xsin⁡x))\left( x+\cos x-\left( x-x\sin x \right) \right)
⇒(x+cos⁡x−x(1−sin⁡x))\Rightarrow \left( x+\cos x-x\left( 1-\sin x \right) \right)
So, cos⁡x+xsin⁡x\cos x+x\sin x can be written as (x+cos⁡x−x(1−sin⁡x))\left( x+\cos x-x\left( 1-\sin x \right) \right)
Hence instead of fraction cos⁡x+xsin⁡xx(x+cos⁡x)\dfrac{\cos x+x\sin x}{x\left( x+\cos x \right)} as (x+cos⁡x)−x(1−sin⁡x)x(x+cos⁡x)\dfrac{\left( x+\cos x \right)-x\left( 1-\sin x \right)}{x\left( x+\cos x \right)}
Now, we will break it into two fractions which are
(x+cos⁡x)x(x+cos⁡x)−x(1−sin⁡x)x(x+cos⁡x)\dfrac{\left( x+\cos x \right)}{x\left( x+\cos x \right)}-\dfrac{x\left( 1-\sin x \right)}{x\left( x+\cos x \right)}
Hence, can be written as
1x−1−sin⁡xx+cos⁡x\dfrac{1}{x}-\dfrac{1-\sin x}{x+\cos x}
Now we will consider this 1x−1−sin⁡xx+cos⁡x\dfrac{1}{x}-\dfrac{1-\sin x}{x+\cos x} to integrate.
Hence,
∫cos⁡x+xsin⁡xx(x+cosx)dx=∫1xdx−∫1−sin⁡xx+cos⁡xdx\int{\dfrac{\cos x+x\sin x}{x\left( x+cosx \right)}dx=}\int{\dfrac{1}{x}dx}-\int{\dfrac{1-\sin x}{x+\cos x}dx}
Now as we know that ∫1xdx=ln⁡(x)+c1\int{\dfrac{1}{x}dx}=\ln \left( x \right)+{{c}_{1}}
So will write in next line and for 1−sin⁡xx+cos⁡x\dfrac{1-\sin x}{x+\cos x} we will take x+cos⁡x=tx+\cos x=t
So, (1−sin⁡x)dx=dt\left( 1-\sin x \right)dx=dt
Hence, in second line we write
=ln⁡(x)+c1−∫dtt=\ln \left( x \right)+{{c}_{1}}-\int{\dfrac{dt}{t}}
So as we know ∫dtt\int{\dfrac{dt}{t}} is ln⁡(t)+c2\ln \left( t \right)+{{c}_{2}}
So,
ln⁡x+c1−∫dtt=ln⁡x+c1−(ln⁡(t)+c2)\ln x+{{c}_{1}}-\int{\dfrac{dt}{t}}=\ln x+{{c}_{1}}-\left( \ln \left( t \right)+{{c}_{2}} \right)
=ln⁡x−ln⁡(t)+c1−c2=\ln x-\ln \left( t \right)+{{c}_{1}}-{{c}_{2}}
Hence, t was (x+cos⁡x)\left( x+\cos x \right) so we will substitute right back and instead c1−c2{{c}_{1}}-{{c}_{2}} we can write simply c to represent constants.
=ln⁡(x)−ln⁡(x+cos⁡x)+c=\ln \left( x \right)-\ln \left( x+\cos x \right)+c
Which can be further written as
ln⁡(xx+cos⁡x)+c\ln \left( \dfrac{x}{x+\cos x} \right)+c
Hence, the answer is ln⁡(xx+cos⁡x)+c\ln \left( \dfrac{x}{x+\cos x} \right)+c.

Note: Students should be careful while breaking the fractions and operating on them simultaneously. Also careful about calculations related to them.