(a) Find: ∫ (2 + sin 2x)/(1 + cos 2x) ex dx OR (b) Evaluate: ∫₀π/4 (1)/(sin x + cos x) dx

Question

\displaystyle\int \dfrac{2 + \sin 2x}{1 + \cos 2x}\, e^x\, dx

(OR)

(b) Evaluate:

\displaystyle\int_{0}^{\pi/4} \dfrac{1}{\sin x + \cos x}\, dx

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Integration Of Exponential

Integration of Exponential Functions

The idea in one line

Integration reverses differentiation. Because the exponential function is the one function that is its own derivative, integrating it is almost as easy as writing it down again.

The base result

Since $\dfrac{d}{dx}\big(e^x\big) = e^x$ , reversing that gives

\int e^x \, dx = e^x + C

That is the whole engine. Every other exponential formula is just this idea adjusted for a coefficient in the exponent or a different base.

When there is a constant in the exponent

For $e^{ax}$ (with $a$ a non-zero constant), differentiating brings a factor of $a$ down. To undo that we must divide by $a$ :

\int e^{ax} \, dx = \frac{e^{ax}}{a} + C

Check it: $\dfrac{d}{dx}\!\left(\dfrac{e^{ax}}{a}\right) = \dfrac{a\,e^{ax}}{a} = e^{ax}.$ ✓ This little "divide by the coefficient of $x$ " step is where most slips happen.

A general base $a^x$

For an exponential with base $a>0,\ a\neq 1$ , recall $\dfrac{d}{dx}\big(a^x\big) = a^x \log a$ . Reversing it, we divide by $\log a$ :

\int a^x \, dx = \frac{a^x}{\log a} + C \qquad (a>0,\ a\neq 1)

When $a=e$ , $\log e = 1$ and this collapses back to $\int e^x\,dx = e^x + C$ — a good consistency check.

Why the $\log a$ appears

Write $a^x = e^{x\log a}$ . Now it is an $e^{kx}$ integral with $k=\log a$ , so $\int a^x dx = \dfrac{e^{x\log a}}{\log a}+C = \dfrac{a^x}{\log a}+C$ . The $\log a$ is exactly the coefficient we divide by. …

Part (b)Concept understanding — Definite Substitution Method

Substitution in Definite Integrals

You already know substitution for indefinite integrals: set $u = g(x)$ , rewrite in terms of $u$ , integrate, then substitute back. For a definite integral there is a cleaner twist — instead of substituting back, you convert the limits of integration to the new variable and finish entirely in $u$ .

Why the limits must change

The limits $a$ and $b$ are $x$ -values. Once you switch to $u = g(x)$ , those numbers no longer describe the start and end of the integration — the corresponding $u$ -values do. Keeping the old numbers would integrate over the wrong interval, like reading a distance in kilometres off a scale marked in miles.

\int_{a}^{b} f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

The steps

Choose $u = g(x)$ , picking something whose derivative already appears in the integrand.
Differentiate: $du = g'(x)\,dx$ .
Convert the limits: the lower limit becomes $u = g(a)$ , the upper becomes $u = g(b)$ .
Integrate in $u$ — no substituting back needed.

Example. Evaluate $\int_0^2 2x\,(x^2+1)^3\,dx$ .

Let $u = x^2 + 1$ , so $du = 2x\,dx$ . When $x=0$ , $u=1$ ; when $x=2$ , $u=5$ . Then

\int_0^2 (x^2+1)^3\,(2x\,dx) = \int_1^5 u^3\,du = \left[\tfrac{u^4}{4}\right]_1^5 = \tfrac{625-1}{4} = 156.

We never returned to $x$ — the converted limits carried the work. …

Start your 14-day free trial to unlock the full solution →

Part (a): simplifying with double-angle identities gives $\int e^x(\tan x+\sec^2x)\,dx=e^x\tan x+C$ . Part (b): writing $\sin x+\cos x=\sqrt2\sin(x+\tfrac\pi4)$ reduces the integral to $\dfrac{1}{\sqrt2}\ln(\sqrt2+1)$ .

Part (a)

The presence of $e^x$ times a sum suggests the pattern $\displaystyle\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+C$ .

1. Simplify the rational part using $1+\cos2x=2\cos^2x$ and $\sin2x=2\sin x\cos x$ :

\frac{2+\sin2x}{1+\cos2x}=\frac{2+2\sin x\cos x}{2\cos^2x}=\frac{1}{\cos^2x}+\frac{\sin x\cos x}{\cos^2x}=\sec^2x+\tan x.

2. Recognise the pattern. The integral becomes

\int e^x(\tan x+\sec^2x)\,dx,

and since $\dfrac{d}{dx}(\tan x)=\sec^2x$ , take $f(x)=\tan x$ , $f'(x)=\sec^2x$ .

\displaystyle\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+C.

3. Apply it: …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 28 on this concept.

CBSE 2026Set 65/3/11 markMCQ
$\int \frac{dx}{2^x + 2^{-x}}$
›Reveal solutionSolution

Rewrite the denominator as $2(2^{2x} + 1)/2^x$ , substitute $u = 2^x$ so $dx = \frac{du}{u \ln 2}$ , and recognize the arctangent integral form. The answer is $\frac{\tan^{-1}(2^x)}{\log 2} + C$ .

The key insight is to transform this exponential expression into a rational function that reveals an arctangent structure. The denominator $2^x + 2^{-x}$ looks symmetric, which suggests we can exploit the relationship between $2^x$ and $2^{-x}$ .

Start by rewriting the denominator in a more workable form. Multiply numerator and denominator by $2^x$ :
$\frac{1}{2^x + 2^{-x}} = \frac{2^x}{2^{2x} + 1}$
So our integral becomes:
$\int \frac{2^x \, dx}{2^{2x} + 1}$
Now the substitution becomes natural. Let $u = 2^x$ . Then:
$\frac{du}{dx} = 2^x \ln 2 = u \ln 2$
which gives us $dx = \frac{du}{u \ln 2}$ .

Substituting into the integral:
$\int \frac{u}{u^2 + 1} \cdot \frac{du}{u \ln 2} = \int \frac{1}{(u^2 + 1) \ln 2} \, du$
Factor out the constant:
$\frac{1}{\ln 2} \int \frac{du}{u^2 + 1}$
This is the standard arctangent integral. We know that $\int \frac{du}{u^2 + 1} = \tan^{-1}(u) + C$ .

Therefore: …
Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\int_1^e\frac{(\log x)^2}{x}\,dx =$
(a) $\frac{1}{3}$
(b) $\frac{1}{3}e^3$
(c) $\frac{1}{3}(e^3 - 1)$
(d) $e^3$
›Reveal solutionSolution

Substitute $t=\log x$ : the integral becomes $\int_0^1 t^2\,dt=\tfrac13$ .

Let $t=\log x$ , so $dt=\tfrac{dx}{x}$ . Limits: $x=1\to t=0$ , $x=e\to t=1$ . Then

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\int_0^{\pi/4}\frac{e^{\tan x}}{\cos^2 x}\,dx =$
(a) $e - 1$
(b) $e + 1$
(c) $\frac{1}{e} + 1$
(d) $\frac{1}{e} - 1$
›Reveal solutionSolution

Substitute $t=\tan x$ (so $dt=\sec^2 x\,dx$ ): the integral becomes $\int_0^1 e^t\,dt=e-1$ .

Let $t=\tan x$ . Then $dt=\sec^2 x\,dx=\dfrac{dx}{\cos^2 x}$ . Limits: $x=0\to t=0$ , $x=\tfrac{\pi}{4}\to t=1$ . So

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\int_0^1\frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx =$
(a) $2(e - 1)$
(b) $e - 1$
(c) $2(e + 1)$
(d) $e + 1$
›Reveal solutionSolution

Substitute $t=\sqrt{x}$ ; the integral becomes $2\int_0^1 e^t\,dt=2(e-1)$ .

Let $t=\sqrt{x}$ , so $dt=\dfrac{dx}{2\sqrt{x}}$ , i.e. $\dfrac{dx}{\sqrt{x}}=2\,dt$ . Limits: $x=0\to t=0$ , $x=1\to t=1$ . Then

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set A1 markMCQ
$\int_0^a\sqrt{a^2 - x^2}\,dx =$
(a) $\frac{\pi}{4}$
(b) $\frac{a^2}{4}$
(c) $\frac{\pi a^2}{4}$
(d) $\pi$
›Reveal solutionSolution

$\int_0^a\sqrt{a^2-x^2}\,dx$ is a quarter-circle area $=\tfrac{\pi a^2}{4}$ .

Using the standard formula $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$ , evaluate from $0$ to $a$ :

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int \frac{dx}{e^{-x}} =$
(a) $\frac{-1}{e^{-x}} + k$
(b) $e^x + k$
(c) $\frac{1}{e^{-x}} \cdot \frac{1}{x^2} + k$
(d) $-e^{-x} + k$
›Reveal solutionSolution

Rewrite $\frac{1}{e^{-x}}$ as $e^{x}$ and integrate; result $e^x+k$ .

Using $\frac{1}{e^{-x}}=e^{x}$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{1} \frac{4\tan^{-1}x}{1 + x^2}\,dx =$
(a) $\frac{\pi^2}{4}$
(b) $\frac{\pi^2}{8}$
(c) $\frac{\pi}{4}$
(d) $\frac{\pi}{8}$
›Reveal solutionSolution

Substitute $u=\tan^{-1}x$ ; the integral becomes $4\int_0^{\pi/4}u\,du=\frac{\pi^2}{8}$ .

Let $u=\tan^{-1}x$ , so $du=\dfrac{dx}{1+x^2}$ . When $x=0,\ u=0$ ; when $x=1,\ u=\frac{\pi}{4}$ . Then …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int e^3 \cdot e^x\,dx =$
(a) $e^x + k$
(b) $\frac{e^{3+x}}{3} + k$
(c) $e^{x+3} + k$
(d) $3e^{x+3} + k$
›Reveal solutionSolution

$e^3$ is a constant multiplier; $\int e^3 e^x\,dx = e^3 e^x + k = e^{x+3}+k$ .

Treat $e^3$ as a constant and pull it out: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int 2^{x+1}\,dx =$
(a) $\frac{2^{x+1}}{\log 2} + k$
(b) $2^{x+1} \cdot \log 2 + k$
(c) $(x + 1)2^x + k$
(d) $2^{x+1} + k$
›Reveal solutionSolution

$\int 2^{x+1}\,dx = \frac{2^{x+1}}{\log 2} + k$ .

Write $2^{x+1} = 2\cdot2^{x}$ and use $\int a^x\,dx = \frac{a^x}{\log a} + k$ with $a = 2$ : …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{2} e^x\,dx =$
(a) $e^2$
(b) $e^2 - 2$
(c) $e^2 - 1$
(d) $e - 1$
›Reveal solutionSolution

$\int_0^2 e^x\,dx = [e^x]_0^2 = e^2 - 1$ .

The antiderivative of $e^x$ is $e^x$ . Applying the limits: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{a} \frac{x\,dx}{2\sqrt{a^2 - x^2}} =$
(a) $\frac{a^2}{2}$
(b) $\frac{a}{2}$
(c) $\frac{a}{4}$
(d) $a$
›Reveal solutionSolution

With $u=a^2-x^2$ , $\int_0^a \frac{x\,dx}{2\sqrt{a^2-x^2}} = \frac12\big[-\sqrt{a^2-x^2}\big]_0^a = \frac{a}{2}$ .

Let $u = a^2 - x^2$ , so $du = -2x\,dx$ , i.e. $x\,dx = -\tfrac12 du$ . Also $\int \frac{x\,dx}{\sqrt{a^2-x^2}} = -\sqrt{a^2-x^2}$ . Therefore

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set E1 markMCQ
$\int_{0}^{1} e^x\,dx =$
(a) $e$
(b) $1 - e$
(c) $e - 1$
(d) $0$
›Reveal solutionSolution

$\int_0^1 e^x\,dx = [e^x]_0^1 = e^1 - e^0 = e - 1$ .

The antiderivative of $e^x$ is $e^x$ . Evaluating:

…

Start your 14-day free trial to see the full solution →

Q.(a) Find: ∫2+sin⁡2x1+cos⁡2x ex dx\displaystyle\int \dfrac{2 + \sin 2x}{1 + \cos 2x}\, e^x\, dx∫1+cos2x2+sin2x​exdx

Part (a)

🎓Unlock everything free for 14 days

Q.(a) Find: $\displaystyle\int \dfrac{2 + \sin 2x}{1 + \cos 2x}\, e^x\, dx$

Unlock everything free for 14 days