Explanation for the correct answer:Finding the value of the given integral: I=∫1+cot^2 x/1+cot xdx (∵ 1+cot^2 x= Cosec^2 )0ex0ex =∫Cosec^2 x/1+cot xdxLet ...

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General Solution of Trigonometric Equation

∫1+cot2x/1+co...

Question

$\int \frac{(1 + c o t^{2} x)}{(1 + c o t x)} d x =$

$- \log |1 + c o t x| + c$

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$\log |1 + \tan x| + c$

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$\log (1 + c o t x) + c$

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$\log (c o t x) + c$

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Solution

The correct option is A
$- \log |1 + c o t x| + c$

Explanation for the correct answer:
Finding the value of the given integral:
$I = \int \frac{1 + c o t^{2} x}{1 + c o t x} d x (∵ 1 + c o t^{2} x = C o s e c^{2}) = \int \frac{C o s e c^{2} x}{1 + c o t x} d x$
Let's consider, $z = 1 + c o t x$
$\Rightarrow d z = 0 - \cos e c^{2} x d x = - c o s e c^{2} x$
Substituting the values $z$ and $d z$ in $I$
$I=- \int \frac{d z}{z} = - \log |z| + c [∵ \int \frac{1}{x} d x = \log |x| + c] = - \log |(1 + c o t x)| + c [∵ z = 1 + c o t x]$
Hence, option (A) is the correct answer.

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Q. Assertion :

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{(1 + a_{1} b_{1}) (1 + a_{1}^{2} b_{1}^{2} - a_{1} b_{1})}{1 + a_{1} b_{1}} & \frac{(1 + a_{1} b_{2}) (1 + a_{1}^{2} b_{2}^{2} - a_{1} b_{2})}{1 + a_{1} b_{2}} & \frac{(1 + a_{1} b_{3}) (1 + a_{1}^{2} b_{3}^{2} - a_{1} b_{3})}{1 + a_{1} b_{3}} \frac{(1 + a_{2} b_{1}) (1 + a_{2}^{2} b_{1}^{2} - a_{2} b_{1})}{1 + a_{2} b_{1}} & \frac{(1 + a_{2} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{2} b_{2})}{1 + a_{2} b_{2}} & \frac{(1 + a_{2} b_{3}) (1 + a_{2}^{2} b_{3}^{2} - a_{2} b_{3})}{1 + a_{2} b_{3}} \frac{(1 + a_{3} b_{1}) (1 + a_{3}^{2} b_{1}^{2} - a_{3} b_{1})}{1 + a_{3} b_{1}} & \frac{(1 + a_{3} b_{2}) (1 + a_{2}^{2} b_{2}^{2} - a_{3} b_{2})}{1 + a_{3} b_{2}} & \frac{(1 + a_{3} b_{3}) (1 + a_{3}^{2} b_{3}^{2} - a_{3} b_{3})}{1 + a_{3} b_{3}} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Δ = 0

Reason:

Δ

can be written as product of two determinants.

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∫(1+cot2x)(1+cotx)dx=

$\int \frac{(1 + c o t^{2} x)}{(1 + c o t x)} d x =$