_{-3/4}^{π-6/8} ((4x+3))\,dx= [TG EAPCET 9 May 2024 Shift 2]

Correct Answer is : −π8log⁡2-π/8 2−8πlog2

Correct Answer is : −π8log⁡2-π/8 2−8πlog2

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TS EAMCET 9 May 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

\int\limits_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log(\sin(4x+3))\,dx=

Solution:

\text{Let } I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log(\sin(4x+3))\,dx \dots \text{(i)}

\Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log\left(\sin\left(4\left(\frac{\pi-6}{8}+\left(-\frac{3}{4}\right)-x\right)+3\right)\right)\,dx

\Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log\left(\sin\left(\frac{\pi}{2}-(4x+3)\right)\right)\,dx

\Rightarrow I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log(\cos(4x+3))\,dx \dots \text{(ii)}.

On adding Eqs. (i) and (ii), we get

2I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log(\sin(4x+3)\cos(4x+3))\,dx

\Rightarrow 2I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log\left(\frac{\sin(2(4x+3))}{2}\right)\,dx

\Rightarrow 2I=\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} \log(\sin(2(4x+3)))\,dx

-\int_{-\frac{3}{4}}^{\frac{\pi-6}{8}} (\log 2)\,dx

\Rightarrow 2I=I-(\log 2)\left[\frac{\pi-6}{8}-\left(-\frac{3}{4}\right)\right]

\Rightarrow I=-\frac{\pi}{8}\log 2

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