ex(tan x+logsec x)

Exercise 11.9 · Q1

e^{x}(\tan x+\log\sec x)

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Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$ , $x\cos x$ , $\log x$ , $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$ : the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

Inverse trig ( $\sin^{-1}x$ ), Logarithmic ( $\log x$ ), Algebraic ( $x^2$ ), Trigonometric ( $\sin x$ ), Exponential ( $e^x$ ).

Whatever comes first in ILATE becomes $u$ ; the rest is $dv$ . This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$ : algebraic before exponential, so $u=x$ , $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ :

\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.

Tip

A single log or a single inverse-trig function ( $\int\log x\,dx$ , $\int\sin^{-1}x\,dx$ ) is still "by parts" — take the other factor as $1$ . And for the special form $\int e^x\big(f(x)+f'(x)\big)dx$ , the answer is simply $e^x f(x)+C$ .

Note

If applying the formula gives you back a multiple of the original integral (as with $\int e^x\sin x\,dx$ ), don't panic — solve for the integral algebraically.

Integration by Parts is one of the most tested methods in the NCERT Class 12 Mathematics chapter on Integrals, and "integration by parts formula ILATE rule" along with "integration by parts class 12 important questions" are among the top searches for students preparing for CBSE board exams and JEE Main calculus. The same ILATE-based technique extends naturally into JEE Advanced integral calculus problems built on this NCERT Class 12 foundation.

Recognising $\tan x$ as the derivative of $\log\sec x$ makes this a direct application of Result 11.2.

Step 1. Identify $f(x)$ and $f'(x)$ . Take $f(x)=\log\sec x$ ; then $f'(x)=\dfrac1{\sec x}\cdot\sec x\tan x=\tan x$ .

Step 2. Match the pattern. $e^x(\tan x+\log\sec x)=e^x[f(x)+f'(x)]$ with $f(x)=\log\sec x$ .

Step 3. Apply Result 11.2. $\displaystyle\int e^x[f(x)+f'(x)]dx=e^xf(x)+c=e^x\log\sec x+c$ .

Step 4. Check. $\dfrac d{dx}[e^x\log\sec x]=e^x\log\sec x+e^x\tan x=e^x(\log\sec x+\tan x)$ , matching the integrand.

✓Final answer

e^x\log\sec x+c

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Q. $e^{x}(\tan x+\log\sec x)$

Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Step 1 — Integration by parts

Step 2 — Complete the integral

Step 3 — Rewrite in the option's form

Q.ex(tan⁡x+log⁡sec⁡x)e^{x}(\tan x+\log\sec x)ex(tanx+logsecx)

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Q. $e^{x}(\tan x+\log\sec x)$

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