Welcome, future engineer. Today, we stand before a problem that, at first glance, might make your heart skip a beat. We are looking at the integral:
It is a complex, trigonometric, algebraic beast. But remember, in the arena of JEE Advanced, intimidation is often just a mask for elegance. Let us peel back the layers of this problem together.
The Detective Work
Analyzing the Denominator
Whenever you see a complex algebraic and trigonometric mix in the denominator, your first instinct should be to investigate it. Do not rush to integrate; instead, pause.
Look at the denominator f(x)=xsinx+cosx. Let us differentiate it with respect to x:
f′(x)=dxd(xsinx+cosx)=(1⋅sinx+x⋅cosx)−sinx=xcosx
The sinx terms cancel out perfectly, leaving us with xcosx. This is our golden key—the hidden structure that will unlock the entire problem.
The Strategic Manipulation
Creating the Derivative
Now that we know the derivative of the denominator is xcosx, we look back at our numerator, which is x2. We need to create xcosx in the numerator to make the integration possible.
We split x2 into x⋅x and strategically multiply and divide the entire expression by cosx:
I=∫xsecx⋅(xsinx+cosx)2xcosxdx
This brilliant manipulation transforms our integral into a product of two distinct functions, setting the stage perfectly for Integration by Parts.
The Integration by Parts Dance
We apply the formula ∫udv=uv−∫vdu. We choose u=xsecx because it is relatively easy to differentiate.
That leaves the rest of the expression as our second part:
We chose this as dv specifically because we engineered its numerator to be the exact derivative of the term inside the denominator's square.
First, we find du by differentiating u=xsecx:
du=(1⋅secx+x⋅secxtanx)dx=secx(1+xtanx)dx
Next, we find v by integrating dv. Substituting t=xsinx+cosx, we get dt=xcosxdx:
v=∫t21dt=−t1=−xsinx+cosx1
The Beautiful Cancellation
Now, we piece everything together using the formula uv−∫vdu:
I=−xsinx+cosxxsecx+∫xsinx+cosxsecx(1+xtanx)dx
Inside the integral, we convert to basic sines and cosines:
secx(1+xtanx)=cosx1+cos2xxsinx=cos2xcosx+xsinx
Notice that the numerator contains the exact same expression as our main denominator: xsinx+cosx. These two terms cancel each other out completely:
∫cos2x(xsinx+cosx)cosx+xsinxdx=∫cos2x1dx=∫sec2xdx=tanx
Conclusion
Finally, we combine all our pieces to write the final answer:
I=tanx−xsinx+cosxxsecx+C
The key takeaway here is to always look for hidden derivatives. By manipulating the numerator to match the derivative of the denominator, we unlocked the entire problem. Keep practicing this technique, and you will find that even the most intimidating integrals are just puzzles waiting to be solved.