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Q2:
Integrate the functions in Exercises 1 to 22. 2. $x \sin 3x$

Solution :

Step 1: Identifying the Integration Technique

To evaluate the integral $I = \int x \sin 3x \, dx$, we employ the method of Integration by Parts. This method is defined by the formula:

$\int u(x) v(x) \, dx = u(x) \int v(x) \, dx - \int \left( \frac{du}{dx} \int v(x) \, dx \right) \, dx$ [Per the Integration by Parts Rule]

Step 2: Selecting Functions (ILATE Rule)

Following the ILATE hierarchy (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), we assign the functions:

  • $u = x$ (Algebraic function)
  • $dv = \sin 3x \, dx$ (Trigonometric function)

Calculating the derivatives and integrals:

  • $\frac{du}{dx} = 1 \implies du = dx$
  • $v = \int \sin 3x \, dx = -\frac{\cos 3x}{3}$ [Per the integration rule $\int \sin(ax) \, dx = -\frac{\cos(ax)}{a} + C$]

Step 3: Executing the Integration by Parts Formula

Substitute $u, v, du,$ and $dv$ into the formula:

$I = x \left( -\frac{\cos 3x}{3} \right) - \int \left( 1 \cdot -\frac{\cos 3x}{3} \right) \, dx$

$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \int \cos 3x \, dx$

Step 4: Final Evaluation of the Integral

Integrate $\cos 3x$:

$\int \cos 3x \, dx = \frac{\sin 3x}{3} + C$ [Per the integration rule $\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$]

Substituting this back into the expression for $I$:

$I = -\frac{x \cos 3x}{3} + \frac{1}{3} \left( \frac{\sin 3x}{3} \right) + C$

$I = -\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$

Visual Representation of the Integrand Area

x y f(x) = x sin(3x)

Final Solution: $-\frac{x \cos 3x}{3} + \frac{\sin 3x}{9} + C$


More Questions from Class 12 Mathematics Integrals EXERCISE 7.6


CBSE Solutions for Class 12 Mathematics Integrals


Chapters in CBSE - Class 12 Mathematics


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