Explanation for the correct answer:Integrating the given integral:Let I=∫1/1+cosx+sinxdxWe know that cos(x)=1 tan^2(x/2)/1+tan^2(x/2), sin(x)=2tan(x/2)/1+ta ...

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Expansions to Remove Indeterminate Form

∫1/1+cosx+sin...

Question

$\int \frac{1}{1 + \cos x + \sin x} d x =$

$(\frac{1}{4}) \log |1 + \tan (\frac{x}{2})| + c$

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$(\frac{1}{2}) \log |1 + \tan (\frac{x}{2})| + c$

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$2 \log |1 + \tan (\frac{x}{2})| + c$

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$(\frac{1}{2}) \log |1 - \tan (\frac{x}{2})| + c$

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Solution

The correct option is A
$(\frac{1}{4}) \log |1 + \tan (\frac{x}{2})| + c$

Explanation for the correct answer:
Integrating the given integral:
Let $I = \int \frac{1}{1 + \cos x + \sin x} d x$
We know that $\cos (x) = \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}, \sin (x) = \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}$
$I = \int \frac{1}{1 + \frac{1 - \tan^{2} (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})} + \frac{2 \tan (\frac{x}{2})}{1 + \tan^{2} (\frac{x}{2})}} d x = \int \frac{\sec^{2} (\frac{x}{2})}{2 (1 + \tan (\frac{x}{2}))} d x$
Substituting
$\tan (\frac{x}{2}) = t Differentiating \frac{1}{2} \sec^{2} (\frac{x}{2}) d x = d t$
Therefore,
$I = \int \frac{d t}{4 (1 + t)} = \frac{1}{4} \log | 1 + t | + c = \frac{1}{4} \log |1 + \tan (\frac{x}{2})| + c$
Hence, $\int \frac{1}{1 + \cos x + \sin x} d x =$ $(\frac{1}{4}) \log | 1 + \tan (\frac{x}{2}) | + c$
Therefore, the correct answer is option (A).

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\int \sqrt{\frac{1}{x - 2}} d x

Q. Assertion :

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Δ = 0

Reason:

Δ

can be written as product of two determinants.

Q. Consider the sequence

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a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}

Let

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then find the value of

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$l o g_{e} (n + 1) - l o g_{e} (n - 1) = 4 a [(\frac{1}{n}) + (\frac{1}{3 n^{3}}) + (\frac{1}{5 n^{5}}) + . . . \infty]$ Find $8 a$ .

Q. If

α

β

γ

are the roots of

x^{3} + a x^{2} + b = 0

b \neq 0

then the determinant

Δ

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Δ = ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{α} & \frac{1}{β} & \frac{1}{γ} \frac{1}{β} & \frac{1}{γ} & \frac{1}{α} \frac{1}{γ} & \frac{1}{α} & \frac{1}{β} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

equals

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