If -01d x-1-x-1-x-2 can be expressed in the form a - b -c-1- where a - b - c are prime numbers- Then value of a - b - c is

Put x = sin2t ⇒ dx = 2cos2t dt I =∫_0^π/42(2 cos^2 t 1)dt/cos t +sin t +cos t sin t +2 =∫_0^π/42 cos^2 t 1/cos t+1dt =∫_0^π/41 2(1 cos^2 t)/(1+cos t)dt =2 ∫_ ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

If ∫01d x/√1+...

Question

If $\int_{0}^{1} \frac{d x}{\sqrt{1 + x} + \sqrt{1 - x} + 2}$ can be expressed in the form $a \sqrt{b} - \frac{π}{c} - 1$ , where a, b, c are prime numbers. Then value of a + b + c is ___

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Solution

Put x = sin2t
$\Rightarrow d x = 2 c o s 2 t d t I = \int_{0}^{\frac{π}{4}} \frac{2 (2 c o s^{2} t - 1) d t}{c o s t + s i n t + c o s t - s i n t + 2} = \int_{0}^{\frac{π}{4}} \frac{2 c o s^{2} t - 1}{c o s t + 1} d t = \int_{0}^{\frac{π}{4}} \frac{1 - 2 (1 - c o s^{2} t)}{(1 + c o s t)} d t$
$= 2 \int_{0}^{\frac{π}{4}} (c o s t - 1) d t + \int_{0}^{\frac{π}{4}} \frac{1}{2 c o s^{2} \frac{t}{2}} d t = 2 [s i n t - t]_{0}^{\frac{π}{4}} + [\frac{1}{2} .2 t a n \frac{t}{2}]_{0}^{\frac{π}{4}} = 2 [\frac{1}{\sqrt{2}} - \frac{π}{4}] + t a n \frac{π}{8} = \sqrt{2} - \frac{π}{2} + \sqrt{2} - 1 = 2 \sqrt{2} - \frac{π}{2} - 1 \Rightarrow a = b = c = 2 S o a + b + c = 6$

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If ∫10dx√1+x+√1−x+2 can be expressed in the form a√b−πc−1, where a, b, c are prime numbers. Then value of a + b + c is ___

Put x = sin2t ⇒dx=2cos2tdtI=∫π402(2cos2t−1)dtcost+sint+cost−sint+2=∫π402cos2t−1cost+1dt=∫π401−2(1−cos2t)(1+cost)dt =2∫π40(cost−1)dt+∫π4012cos2t2dt=2[sint−t]π40+[12.2tant2]π40=2[1√2−π4]+tanπ8=√2−π2+√2−1=2√2−π2−1⇒a=b=c=2So a+b+c=6

If $\int_{0}^{1} \frac{d x}{\sqrt{1 + x} + \sqrt{1 - x} + 2}$ can be expressed in the form $a \sqrt{b} - \frac{π}{c} - 1$ , where a, b, c are prime numbers. Then value of a + b + c is ___