NCERT Class 12 Maths Integrals Exercise 7.6 Guide

Understanding Integration by Parts

Integration by parts is derived directly from the Product Rule of Differentiation. Let $u$ and $v$ be two differentiable functions of a single variable $x$. The product rule states that the derivative of their product is $d/dx(uv) = u(dv/dx) + v(du/dx)$. Integrating both sides with respect to $x$ and rearranging the terms gives us the standard Integration by Parts formula:

$\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx$

Here, $u$ is chosen as the 'First Function' and $v$ is chosen as the 'Second Function'. The choice of which function to designate as $u$ and which as $v$ is crucial. Choosing the wrong first function can lead to an integral that is more complex than the original. To systematically solve this, mathematicians created the ILATE rule.

The ILATE Priority Order for Selecting the First Function (u)

I - Inverse Trigonometric Functions

Functions like $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$. These have top priority to be chosen as $u$ because we cannot integrate them directly using basic formulas.

L - Logarithmic Functions

Functions like $\log x$ or $\ln(x^2)$. These are prioritized next. Similar to inverse trig, we don't have a direct integral formula for logarithms, so they must be differentiated.

A - Algebraic Functions

Polynomial terms like $x$, $x^2$, $3x+5$, constant values, or fractional powers. These are easily differentiated to lower their degree, or integrated.

T - Trigonometric Functions

Standard trig functions such as $\sin x$, $\cos x$, $\sec^2 x$. They are highly integration-friendly, making them ideal second functions ($v$).

E - Exponential Functions

Functions like $e^x$, $a^x$. They are the easiest to integrate and always retain their form, hence placed at the lowest priority for $u$ (typically chosen as $v$).

Step-by-Step Guide to Applying Integration by Parts

1. Identify the Functions — Look at the integrand and identify the two multiplied functions. If only one function is given (like $\int \log x \, dx$), write it as a product with $1$ (e.g., $\int (\log x \cdot 1) \, dx$).
2. Apply ILATE Rule — Compare the two functions using the ILATE order. Assign the function appearing earlier in the acronym as the first function $u$ and the other as the second function $v$.
3. Write the Formula — Substitute $u$ and $v$ into the Integration by Parts formula: $u \int v \, dx - \int (u' \int v \, dx) dx$.
4. Integrate & Simplify — Perform the inner integration first, then differentiate $u$, multiply the results, and evaluate the final outer integral. Do not forget to add the constant of integration $C$.

Fully Solved Core Problems

• Example 1: Find $\int x \cos x \, dx$ Step 1: Identify functions. $x$ is algebraic (A), $\cos x$ is trigonometric (T). Using ILATE, Algebraic comes before Trigonometric, so $u = x$ and $v = \cos x$. Step 2: Apply the formula: $= x \int \cos x \, dx - \int \left( \frac{d}{dx}(x) \int \cos x \, dx \right) dx$ Step 3: Evaluate integrals: $= x(\sin x) - \int (1 \cdot \sin x) dx$ $= x \sin x - (-\cos x) + C$ $= x \sin x + \cos x + C$
• Example 2: Find $\int \log x \, dx$ Step 1: Treat this as $\dots \int (\log x \cdot 1) \, dx$. Here, $\log x$ is logarithmic (L) and $1$ is algebraic (A). Therefore, $u = \log x$ and $v = 1$. Step 2: Apply the formula: $= \log x \int 1 \, dx - \int \left( \frac{d}{dx}(\log x) \int 1 \, dx \right) dx$ Step 3: Evaluate integrals: $= \log x \cdot (x) - \int \left( \frac{1}{x} \cdot x \right) dx$ $= x \log x - \int 1 \, dx$ $= x \log x - x + C = x(\log x - 1) + C$
• Example 3: Evaluate $\int e^x (\tan x + \sec^2 x) \, dx$ Step 1: Recognize the special theorem form: $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$. Step 2: Identify $f(x)$. Let $f(x) = \tan x$, then $f'(x) = \sec^2 x$. Step 3: Write down the final solution directly based on the theorem: $= e^x \tan x + C$

Crucial Exam Traps & Board Tips

1. Double Integration-by-Parts: In problems containing polynomials like $x^2 e^x$ or cyclic functions like $e^x \sin x$, you will have to apply the integration-by-parts rule twice. For cyclic integrals, designate the original integral as $I$, apply the rule twice until $I$ reappears on the right-hand side, then solve algebraically for $I$.
2. The Minus Sign Trap: When evaluating $- \int (u' \int v \, dx) dx$, pay extra attention to negatives. If $u'$ or the inner integral produces a negative sign, distribute it carefully across all brackets.
3. Write the Theorem: When using the special formula $\int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C$, explicitly state the identity and identify $f(x)$ and $f'(x)$ to secure full method marks from CBSE evaluators.

Practice Questions with Solutions

• Q: Evaluate $\int x e^x \, dx$ A: Step 1: Using the ILATE rule, $x$ is algebraic (A) and $e^x$ is exponential (E). Thus, first function $u = x$ and second function $v = e^x$. Step 2: Write down the integration by parts formula: $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) dx$. Step 3: Substitute the terms: $= x \int e^x \, dx - \int \left( \frac{d}{dx}(x) \int e^x \, dx \right) dx$ Step 4: Compute the integration: $= x e^x - \int (1 \cdot e^x) dx$ $= x e^x - e^x + C$ Final answer: $e^x(x - 1) + C$
• Q: Find the integral of $\int x^2 \log x \, dx$ A: Step 1: Under the ILATE order, Logarithmic (L) comes before Algebraic (A). Therefore, let $u = \log x$ and $v = x^2$. Step 2: Apply the integration by parts formula: $= \log x \int x^2 \, dx - \int \left( \frac{d}{dx}(\log x) \int x^2 \, dx \right) dx$ Step 3: Integrate and simplify terms: $= \log x \left( \frac{x^3}{3} \right) - \int \left( \frac{1}{x} \cdot \frac{x^3}{3} \right) dx$ $= \frac{x^3 \log x}{3} - \frac{1}{3} \int x^2 \, dx$ Step 4: Compute the final integral: $= \frac{x^3 \log x}{3} - \frac{1}{3} \left( \frac{x^3}{3} \right) + C$ $= \frac{x^3 \log x}{3} - \frac{x^3}{9} + C$ Final answer: $\frac{x^3}{3}\left( \log x - \frac{1}{3} \right) + C$
• Q: Evaluate $\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx$ A: Step 1: Compare this integral to the standard form $\int e^x [f(x) + f'(x)] dx$. Step 2: Let $f(x) = \frac{1}{x} = x^{-1}$. Then, differentiate to find $f'(x)$: $f'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}$. Step 3: Since the integrand perfectly matches $e^x [f(x) + f'(x)]$, apply the identity directly: $\int e^x \left(\frac{1}{x} + \left(-\frac{1}{x^2}\right)\right) dx = e^x f(x) + C$ Final answer: $\frac{e^x}{x} + C$
• Q: Evaluate $\int \sin^{-1}x \, dx$ A: Step 1: Set up the integral with a dummy function: $\int (\sin^{-1}x \cdot 1) dx$. Here, $u = \sin^{-1}x$ (Inverse) and $v = 1$ (Algebraic). Step 2: Apply integration by parts: $= \sin^{-1}x \int 1 \, dx - \int \left( \frac{d}{dx}(\sin^{-1}x) \int 1 \, dx \right) dx$ $= x \sin^{-1}x - \int \frac{x}{\sqrt{1-x^2}} \, dx$ Step 3: Solve the remaining integral using substitution. Let $1-x^2 = t \implies -2x \, dx = dt \implies x \, dx = -\frac{dt}{2}$: $\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{-dt}{2\sqrt{t}} = -\sqrt{t} = -\sqrt{1-x^2}$ Step 4: Substitute back into our expression: $= x \sin^{-1}x - (-\sqrt{1-x^2}) + C$ Final answer: $x \sin^{-1}x + \sqrt{1-x^2} + C$

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