Choose the correct answer- e x 1- x -cos 2 e x x d x is equal to a - e - x x - Cllbtan xe - x - Cllctan e - x - Clld - e - x - Cl

∫e^x(1+x)/cos^2 (e^x x)dx Let xe^x =t ⇒ (xe^x +e^x)=dt/dx⇒ dx =dt/e^x (x+1) ∴∫e^x (1+x)/cos^2 (e^x x)dx =∫e^x (1+x)/cos^2 t×dt/e^x(1+x) =∫1/cos^2 tdt =∫ sec^2 ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

Choose the co...

Question

Choose the correct answer.
$\int \frac{e^{x} (1 + x)}{c o s^{2} (e^{x} x)} d x$ is equal to
$(a) - c o t (e^{x} x) + C (b) t a n (x e^{x}) + C (c) t a n (e^{x}) + C (d) c o t (e^{x}) + C$

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Solution

$\int \frac{e^{x} (1 + x)}{c o s^{2} (e^{x} x)} d x$
Let $x e^{x} = t \Rightarrow (x e^{x} + e^{x}) = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{e^{x} (x + 1)}$
$∴ \int \frac{e^{x} (1 + x)}{c o s^{2} (e^{x} x)} d x = \int \frac{e^{x} (1 + x)}{c o s^{2} t} \times \frac{d t}{e^{x} (1 + x)} = \int \frac{1}{c o s^{2} t} d t = \int s e c^{2} t d t = t a n t + C = t a n (x e^{x}) + C$
Hence, option (b)is correct.

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equals

\int \frac{e^{\sqrt{x}}}{\sqrt{x}} . (x + \sqrt{x}) d x

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Choose the correct answer. ∫ex(1+x)cos2(exx)dx is equal to (a)−cot(exx)+C(b)tan(xex)+C(c)tan(ex)+C(d)cot(ex)+C

∫ex(1+x)cos2(exx)dx Let xex=t⇒(xex+ex)=dtdx⇒dx=dtex(x+1) ∴∫ex(1+x)cos2(exx)dx=∫ex(1+x)cos2t×dtex(1+x)=∫1cos2tdt=∫sec2tdt=tant+C=tan(xex)+C Hence, option (b)is correct.

Choose the correct answer.
$\int \frac{e^{x} (1 + x)}{c o s^{2} (e^{x} x)} d x$ is equal to
$(a) - c o t (e^{x} x) + C (b) t a n (x e^{x}) + C (c) t a n (e^{x}) + C (d) c o t (e^{x}) + C$