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Q9:
Integrate the functions in Exercises 1 to 22. 9. $x \cos^{-1} x$

Solution :

Step 1: Application of Integration by Parts

To evaluate the integral $I = \int x \cos^{-1} x \, dx$, we employ the method of Integration by Parts, which states: $\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx$ [Per the Integration by Parts Theorem].

According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we assign:

  • $u = \cos^{-1} x$
  • $dv = x \, dx$

Differentiating $u$ and integrating $dv$ yields:

  • $\frac{du}{dx} = -\frac{1}{\sqrt{1-x^2}}$
  • $v = \int x \, dx = \frac{x^2}{2}$

Step 2: Substitution into the Formula

Applying the components to the integration formula:

$I = \left( \cos^{-1} x \right) \left( \frac{x^2}{2} \right) - \int \left( -\frac{1}{\sqrt{1-x^2}} \right) \left( \frac{x^2}{2} \right) dx$

$I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$

Step 3: Evaluating the Remaining Integral

To integrate $\int \frac{x^2}{\sqrt{1-x^2}} \, dx$, we manipulate the numerator to create a term that matches the denominator:

$\int \frac{x^2}{\sqrt{1-x^2}} \, dx = \int \frac{-(1-x^2) + 1}{\sqrt{1-x^2}} \, dx$

$= -\int \frac{1-x^2}{\sqrt{1-x^2}} \, dx + \int \frac{1}{\sqrt{1-x^2}} \, dx$

$= -\int \sqrt{1-x^2} \, dx + \int \frac{1}{\sqrt{1-x^2}} \, dx$

Step 4: Applying Standard Integral Formulas

We use the following standard results:

  1. $\int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)$
  2. $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right)$

Thus, $\int \frac{x^2}{\sqrt{1-x^2}} \, dx = -\left( \frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right) + \sin^{-1} x$

$= -\frac{x}{2}\sqrt{1-x^2} - \frac{1}{2}\sin^{-1} x + \sin^{-1} x = -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x$

Step 5: Final Assembly

Substitute the result back into the expression for $I$:

$I = \frac{x^2}{2} \cos^{-1} x + \frac{1}{2} \left[ -\frac{x}{2}\sqrt{1-x^2} + \frac{1}{2}\sin^{-1} x \right] + C$

$I = \frac{x^2}{2} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1} x + C$

x y f(x) = x cos⁻¹ x

Final Solution: $\frac{x^2}{2} \cos^{-1} x - \frac{x}{4}\sqrt{1-x^2} + \frac{1}{4}\sin^{-1} x + C$


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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