JEE Mains · Maths · STD 12 - 7.1 indefinite integral
The integral \(\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{4 x^{2}-4 x+6}} d x\) is equal to (where \(c\) is a constant of integration)
- A \(\frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5}+c\)
- B \(\frac{1}{2} \cos \sqrt{(2 x+1)^{2}+5}+c\)
- C \(\frac{1}{2} \cos \sqrt{(2 x-1)^{2}+5}+c\)
- D \(\frac{1}{2} \sin \sqrt{(2 x+1)^{2}+5}+c\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2} \sin \sqrt{(2 x-1)^{2}+5}+c\)
Step-by-step Solution
Detailed explanation
\(\int \frac{(2 x-1) \cos \sqrt{(2 x-1)^{2}+5}}{\sqrt{(2 x-1)^{2}+5}} d x\) \((2 x-1)^{2}+5=t^{2}\) \(2(2 x-1) 2 d x=2 t d t\) \(2 \sqrt{t^{2}-5} d x=t d t\) So \(\int \frac{\sqrt{t^{2}-5} \cos t}{2 \sqrt{t^{2}-5}} d t=\frac{1}{2} \sin t+c\)…
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