JEE Mains · Maths · STD 12 - 7.2 definite integral
Let for \(f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x, \quad \mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x\) and \(\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x\). Then \(7 \mathrm{I}_1+12 \mathrm{I}_2\) is equal to :
- A \(2\)
- B \(1\)
- C \(2 \pi\)
- D \(\pi\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(f(x) =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \) \( =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \) \( =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \) \( =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x \)…
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