Exams Joint Entrance Examination Mathematics Definite Integration Question 31 Question 31 of 246 Medium MCQ
The integral 80 ∫ 0 π 4 ( sin θ + cos θ 9 + 16 sin 2 θ ) d θ 80\int _{0}^{\frac{\pi }{4}}(\frac{\sin \theta +\cos \theta }{9+16\sin 2\theta })d\theta 80 ∫ 0 4 π ( 9 + 16 s i n 2 θ s i n θ + c o s θ ) d θ is equal to :
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\[ I = 80 ∫ 0 π 4 ( sin θ + cos θ 9 + 16 ( 2 sin θ ⋅ cos θ ) ) d θ = 80 ∫ 0 π 4 sin θ + cos θ 9 − 16 ( 1 − 2 sin θ ⋅ cos θ − 1 ) d θ \begin{aligned}
& I=80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16(2 \sin \theta \cdot \cos \theta)}\right) d \theta \\
& =80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9-16(1-2 \sin \theta \cdot \cos \theta-1)} d \theta
\end{aligned} I = 80 ∫ 0 4 π ( 9 + 16 ( 2 sin θ ⋅ cos θ ) sin θ + cos θ ) d θ = 80 ∫ 0 4 π 9 − 16 ( 1 − 2 sin θ ⋅ cos θ − 1 ) sin θ + cos θ d θ \]
\[ = 80 ∫ 0 π 4 sin θ + cos θ 9 + 16 − 16 ( sin θ − cos θ ) 2 d θ Let sin θ − cos θ = t ( cos θ + sin θ ) d θ = d t = 80 ∫ − 1 0 d t 25 − 16 t 2 = 80 16 ∫ − 1 0 d t ( 5 4 ) 2 − t 2 = 5 2 ( 5 4 ) ln ∣ 5 4 + t 5 4 − t ∣ ] − 1 0 = 2 ln ( 1 ) + 4 ln 3 = 4 ln 3 \begin{aligned}
&=80 \int_0^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^2} \mathrm{~d} \theta\\
&\text { Let } \sin \theta-\cos \theta=\mathrm{t}\\
&(\cos \theta+\sin \theta) \mathrm{d} \theta=\mathrm{dt}\\
&=80 \int_{-1}^0 \frac{\mathrm{dt}}{25-16 \mathrm{t}^2}\\
&=\frac{80}{16} \int_{-1}^0 \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^2-\mathrm{t}^2}\\
&\left.=\frac{5}{2\left(\frac{5}{4}\right)} \ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]_{-1}^0\\
&=2 \ln (1)+4 \ln 3\\
&=4 \ln 3
\end{aligned} = 80 ∫ 0 4 π 9 + 16 − 16 ( sin θ − cos θ ) 2 sin θ + cos θ d θ Let sin θ − cos θ = t ( cos θ + sin θ ) d θ = dt = 80 ∫ − 1 0 25 − 16 t 2 dt = 16 80 ∫ − 1 0 ( 4 5 ) 2 − t 2 dt = 2 ( 4 5 ) 5 ln 4 5 − t 4 5 + t ] − 1 0 = 2 ln ( 1 ) + 4 ln 3 = 4 ln 3 \] Answer the question to unlock