Animated Solution for Mathematics - Indefinite Integration: Let $\int x^3\sin xdx=g(x)+C$ where C is the constant of integration. If $8(g(\frac{\pi }{2})+g^{\prime }(\frac{\pi }{2}))=\alpha {\pi }^3+\beta {\pi }^2+\gamma .$ $\alpha ,\beta ,\gamma \in Z$, then $\alpha +\beta -\gamma$ equals :

Mastering Integration by Parts

Welcome, fellow traveler on the path to JEE Advanced mastery. Today, we are tackling an integration problem that, at first glance, looks like a test of endurance.

We are asked to find the function $g(x)$ defined by the integral $\int x^3\sin xdx$ and then evaluate a specific expression involving $g(\frac{\pi }{2})$ and $g^{\prime }(\frac{\pi }{2})$.

Many students see $x^3\sin x$ and immediately start writing out three rounds of integration by parts. While that is a valid path, it is also a path paved with potential for silly arithmetic errors. Today, we learn to be smarter, not just harder.

Your Secret Weapon

When you see a product of a polynomial and a trigonometric function, the Tabular Method (or DI Method) is your best friend. It is a structured, visual way to perform integration by parts repeatedly.

We set up three columns: one for the alternating sign, one for the function to differentiate (D), and one for the function to integrate (I).

We choose $u=x^3$ for the D column because its derivative will eventually become zero, terminating the process. We choose $dv=\sin x$ for the I column because it is easy to integrate. We start with a positive sign, and then we alternate: $(+),(-),(+),(-)$.

Following the steps: 1. Row 1: Sign $(+)$, D term $x^3$, I term $\sin x$. 2. Row 2: Sign $(-)$, D term $3x^2$, I term $-\cos x$. 3. Row 3: Sign $(+)$, D term $6x$, I term $-\sin x$. 4. Row 4: Sign $(-)$, D term $6$, I term $\cos x$. 5. Row 5: Sign $(+)$, D term $0$, I term $\sin x$.

By multiplying diagonally—Sign $\times$ D $\times$ I—we construct our function $g(x)$:

A Moment of Clarity

Now, the problem asks for $g^{\prime }(\frac{\pi }{2})$. A novice would differentiate the entire expression we just found. But you? You know better.

By the Fundamental Theorem of Calculus, the derivative of the integral is simply the integrand itself. Therefore, $g^{\prime }(x)=x^3\sin x$.

Evaluating this at $x=\frac{\pi }{2}$ is trivial:

Now we evaluate $g(\frac{\pi }{2})$. Look at our expression for $g(x)$. Notice that $\cos (\frac{\pi }{2})=0$.

This means the terms $-x^3\cos x$ and $6x\cos x$ vanish completely! We are left with:

Finally, we substitute these into the expression $8(g(\frac{\pi }{2})+g^{\prime }(\frac{\pi }{2}))$:

Comparing this to $\alpha {\pi }^3+\beta {\pi }^2+\gamma$, we find $\alpha =1$, $\beta =6$, and $\gamma =-48$.

The final calculation $\alpha +\beta -\gamma =1+6-(-48)=55$. You have conquered the problem with elegance and precision. Keep this mindset, and no integral will ever stand in your way.