(C) : Let, \(\mathrm{I}=\int 1 \cdot \cot ^{-1} \mathrm{x} d x\) \(I=\left[\cot ^{-1} x \cdot x-\int \frac{-1}{1+x^{2}} x d x\right]\) \(I=x \cot ^{-1} x+\int \frac{x}{1+x^{2}} d x\) Let, \(\quad 1+\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \int \frac{1}{\mathrm{t}} \mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \log |\mathrm{t}|\) \(I=x \cot ^{-1} \mathrm{x}+\frac{1}{2} \log \left|1+\mathrm{x}^{2}\right|\) Now, \(\int_{-1}^{1} \cot ^{-1} x d x=\left[x \cot ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|\right]_{-1}^{1}\) \(=\left[1 \cot ^{-1} 1+\frac{1}{2} \log |2|\right]-\left[(-1) \cot ^{-1}(-1)+\frac{1}{2} \log 2\right]\) \(=\cot ^{-1} 1+\frac{1}{2} \log 2+\cot ^{-1}(-1)-\frac{1}{2} \log 2\) \(=\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{4 \pi}{4}\) \(=\pi\)
GUJCET-2021
Integral Calculus
86749
\(\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=\)
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 0
4 \(-\pi\)
Explanation:
(C): Let, \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x\) Let, \(f(x)=\frac{\sin ^{3}(x) \cdot \cos ^{3}(x)}{\cos ^{2}(x)-\sin ^{2}(x)}\) \(f(-x)=\frac{\sin ^{3}(-x) \cdot \cos ^{3}(-x)}{\cos ^{2}(-x)-\sin ^{2}(-x)}\) \(f(-x)=\frac{(-1)^{3} \sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=\frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=-f(x)\) So, \(\mathrm{f}(\mathrm{x})\) is an odd function. \(\therefore \int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=0\)
GUJCET-2011
Integral Calculus
86750
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\) \(\pi^{2}, \forall \in \mathbf{R}\). Then \(\int_{0}^{\pi} f(x) \sin x d x\) is equal to
1 \(\frac{\pi^{2}}{4}\)
2 \(\frac{\pi^{2}}{2}\)
3 \(2 \pi^{2}\)
4 \(\pi^{2}\)
Explanation:
(D) : Given,\(f(x)+f(\pi-x)=\pi^{2} \forall \in R\) \(I=\int_{0}^{\pi} f(x) \cdot \sin x d x\) \(I=\int_{0}^{\pi} f(\pi-x) \cdot \sin (\pi-x) d x\) \(I=\int_{0}^{\pi} f(\pi-x) \sin x d x \tag{ii}\) On adding equation (i) and (ii) we get- \(2 I=\int_{0}^{\pi}[f(x)+f(\pi-x)] \sin x d x\) \(2 I=\int_{0}^{\pi} \pi^{2} \sin x d x \Rightarrow 2 I=\pi^{2} \int_{0}^{\pi} \sin x d x\) \(2 I=\pi^{2}[-\cos x]_{0}^{\pi} \Rightarrow 2 I=-\pi^{2}(-1-1)\) \(2 I=\pi^{2} \times 2 \Rightarrow I=\pi^{2}\)
JEE Main-2023-06.04.2023
Integral Calculus
86758
\(\int_{0}^{\frac{\pi}{2}} \sin ^{5}\left(\frac{x}{2}\right) \cdot \sin x d x=\)
(C) : Let, \(\mathrm{I}=\int 1 \cdot \cot ^{-1} \mathrm{x} d x\) \(I=\left[\cot ^{-1} x \cdot x-\int \frac{-1}{1+x^{2}} x d x\right]\) \(I=x \cot ^{-1} x+\int \frac{x}{1+x^{2}} d x\) Let, \(\quad 1+\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \int \frac{1}{\mathrm{t}} \mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \log |\mathrm{t}|\) \(I=x \cot ^{-1} \mathrm{x}+\frac{1}{2} \log \left|1+\mathrm{x}^{2}\right|\) Now, \(\int_{-1}^{1} \cot ^{-1} x d x=\left[x \cot ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|\right]_{-1}^{1}\) \(=\left[1 \cot ^{-1} 1+\frac{1}{2} \log |2|\right]-\left[(-1) \cot ^{-1}(-1)+\frac{1}{2} \log 2\right]\) \(=\cot ^{-1} 1+\frac{1}{2} \log 2+\cot ^{-1}(-1)-\frac{1}{2} \log 2\) \(=\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{4 \pi}{4}\) \(=\pi\)
GUJCET-2021
Integral Calculus
86749
\(\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=\)
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 0
4 \(-\pi\)
Explanation:
(C): Let, \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x\) Let, \(f(x)=\frac{\sin ^{3}(x) \cdot \cos ^{3}(x)}{\cos ^{2}(x)-\sin ^{2}(x)}\) \(f(-x)=\frac{\sin ^{3}(-x) \cdot \cos ^{3}(-x)}{\cos ^{2}(-x)-\sin ^{2}(-x)}\) \(f(-x)=\frac{(-1)^{3} \sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=\frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=-f(x)\) So, \(\mathrm{f}(\mathrm{x})\) is an odd function. \(\therefore \int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=0\)
GUJCET-2011
Integral Calculus
86750
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\) \(\pi^{2}, \forall \in \mathbf{R}\). Then \(\int_{0}^{\pi} f(x) \sin x d x\) is equal to
1 \(\frac{\pi^{2}}{4}\)
2 \(\frac{\pi^{2}}{2}\)
3 \(2 \pi^{2}\)
4 \(\pi^{2}\)
Explanation:
(D) : Given,\(f(x)+f(\pi-x)=\pi^{2} \forall \in R\) \(I=\int_{0}^{\pi} f(x) \cdot \sin x d x\) \(I=\int_{0}^{\pi} f(\pi-x) \cdot \sin (\pi-x) d x\) \(I=\int_{0}^{\pi} f(\pi-x) \sin x d x \tag{ii}\) On adding equation (i) and (ii) we get- \(2 I=\int_{0}^{\pi}[f(x)+f(\pi-x)] \sin x d x\) \(2 I=\int_{0}^{\pi} \pi^{2} \sin x d x \Rightarrow 2 I=\pi^{2} \int_{0}^{\pi} \sin x d x\) \(2 I=\pi^{2}[-\cos x]_{0}^{\pi} \Rightarrow 2 I=-\pi^{2}(-1-1)\) \(2 I=\pi^{2} \times 2 \Rightarrow I=\pi^{2}\)
JEE Main-2023-06.04.2023
Integral Calculus
86758
\(\int_{0}^{\frac{\pi}{2}} \sin ^{5}\left(\frac{x}{2}\right) \cdot \sin x d x=\)
(C) : Let, \(\mathrm{I}=\int 1 \cdot \cot ^{-1} \mathrm{x} d x\) \(I=\left[\cot ^{-1} x \cdot x-\int \frac{-1}{1+x^{2}} x d x\right]\) \(I=x \cot ^{-1} x+\int \frac{x}{1+x^{2}} d x\) Let, \(\quad 1+\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \int \frac{1}{\mathrm{t}} \mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \log |\mathrm{t}|\) \(I=x \cot ^{-1} \mathrm{x}+\frac{1}{2} \log \left|1+\mathrm{x}^{2}\right|\) Now, \(\int_{-1}^{1} \cot ^{-1} x d x=\left[x \cot ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|\right]_{-1}^{1}\) \(=\left[1 \cot ^{-1} 1+\frac{1}{2} \log |2|\right]-\left[(-1) \cot ^{-1}(-1)+\frac{1}{2} \log 2\right]\) \(=\cot ^{-1} 1+\frac{1}{2} \log 2+\cot ^{-1}(-1)-\frac{1}{2} \log 2\) \(=\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{4 \pi}{4}\) \(=\pi\)
GUJCET-2021
Integral Calculus
86749
\(\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=\)
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 0
4 \(-\pi\)
Explanation:
(C): Let, \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x\) Let, \(f(x)=\frac{\sin ^{3}(x) \cdot \cos ^{3}(x)}{\cos ^{2}(x)-\sin ^{2}(x)}\) \(f(-x)=\frac{\sin ^{3}(-x) \cdot \cos ^{3}(-x)}{\cos ^{2}(-x)-\sin ^{2}(-x)}\) \(f(-x)=\frac{(-1)^{3} \sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=\frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=-f(x)\) So, \(\mathrm{f}(\mathrm{x})\) is an odd function. \(\therefore \int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=0\)
GUJCET-2011
Integral Calculus
86750
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\) \(\pi^{2}, \forall \in \mathbf{R}\). Then \(\int_{0}^{\pi} f(x) \sin x d x\) is equal to
1 \(\frac{\pi^{2}}{4}\)
2 \(\frac{\pi^{2}}{2}\)
3 \(2 \pi^{2}\)
4 \(\pi^{2}\)
Explanation:
(D) : Given,\(f(x)+f(\pi-x)=\pi^{2} \forall \in R\) \(I=\int_{0}^{\pi} f(x) \cdot \sin x d x\) \(I=\int_{0}^{\pi} f(\pi-x) \cdot \sin (\pi-x) d x\) \(I=\int_{0}^{\pi} f(\pi-x) \sin x d x \tag{ii}\) On adding equation (i) and (ii) we get- \(2 I=\int_{0}^{\pi}[f(x)+f(\pi-x)] \sin x d x\) \(2 I=\int_{0}^{\pi} \pi^{2} \sin x d x \Rightarrow 2 I=\pi^{2} \int_{0}^{\pi} \sin x d x\) \(2 I=\pi^{2}[-\cos x]_{0}^{\pi} \Rightarrow 2 I=-\pi^{2}(-1-1)\) \(2 I=\pi^{2} \times 2 \Rightarrow I=\pi^{2}\)
JEE Main-2023-06.04.2023
Integral Calculus
86758
\(\int_{0}^{\frac{\pi}{2}} \sin ^{5}\left(\frac{x}{2}\right) \cdot \sin x d x=\)
(C) : Let, \(\mathrm{I}=\int 1 \cdot \cot ^{-1} \mathrm{x} d x\) \(I=\left[\cot ^{-1} x \cdot x-\int \frac{-1}{1+x^{2}} x d x\right]\) \(I=x \cot ^{-1} x+\int \frac{x}{1+x^{2}} d x\) Let, \(\quad 1+\mathrm{x}^{2}=\mathrm{t}\) \(2 \mathrm{xdx}=\mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \int \frac{1}{\mathrm{t}} \mathrm{dt}\) \(\mathrm{I}=\mathrm{x} \cot ^{-1} \mathrm{x}+\frac{1}{2} \log |\mathrm{t}|\) \(I=x \cot ^{-1} \mathrm{x}+\frac{1}{2} \log \left|1+\mathrm{x}^{2}\right|\) Now, \(\int_{-1}^{1} \cot ^{-1} x d x=\left[x \cot ^{-1} x+\frac{1}{2} \log \left|1+x^{2}\right|\right]_{-1}^{1}\) \(=\left[1 \cot ^{-1} 1+\frac{1}{2} \log |2|\right]-\left[(-1) \cot ^{-1}(-1)+\frac{1}{2} \log 2\right]\) \(=\cot ^{-1} 1+\frac{1}{2} \log 2+\cot ^{-1}(-1)-\frac{1}{2} \log 2\) \(=\frac{\pi}{4}+\frac{3 \pi}{4}=\frac{4 \pi}{4}\) \(=\pi\)
GUJCET-2021
Integral Calculus
86749
\(\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=\)
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{4}\)
3 0
4 \(-\pi\)
Explanation:
(C): Let, \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x\) Let, \(f(x)=\frac{\sin ^{3}(x) \cdot \cos ^{3}(x)}{\cos ^{2}(x)-\sin ^{2}(x)}\) \(f(-x)=\frac{\sin ^{3}(-x) \cdot \cos ^{3}(-x)}{\cos ^{2}(-x)-\sin ^{2}(-x)}\) \(f(-x)=\frac{(-1)^{3} \sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=\frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x}\) \(f(-x)=-f(x)\) So, \(\mathrm{f}(\mathrm{x})\) is an odd function. \(\therefore \int_{-\pi / 2}^{\pi / 2} \frac{\sin ^{3} x \cdot \cos ^{3} x}{\cos ^{2} x-\sin ^{2} x} d x=0\)
GUJCET-2011
Integral Calculus
86750
Let \(f(x)\) be a function satisfying \(f(x)+f(\pi-x)=\) \(\pi^{2}, \forall \in \mathbf{R}\). Then \(\int_{0}^{\pi} f(x) \sin x d x\) is equal to
1 \(\frac{\pi^{2}}{4}\)
2 \(\frac{\pi^{2}}{2}\)
3 \(2 \pi^{2}\)
4 \(\pi^{2}\)
Explanation:
(D) : Given,\(f(x)+f(\pi-x)=\pi^{2} \forall \in R\) \(I=\int_{0}^{\pi} f(x) \cdot \sin x d x\) \(I=\int_{0}^{\pi} f(\pi-x) \cdot \sin (\pi-x) d x\) \(I=\int_{0}^{\pi} f(\pi-x) \sin x d x \tag{ii}\) On adding equation (i) and (ii) we get- \(2 I=\int_{0}^{\pi}[f(x)+f(\pi-x)] \sin x d x\) \(2 I=\int_{0}^{\pi} \pi^{2} \sin x d x \Rightarrow 2 I=\pi^{2} \int_{0}^{\pi} \sin x d x\) \(2 I=\pi^{2}[-\cos x]_{0}^{\pi} \Rightarrow 2 I=-\pi^{2}(-1-1)\) \(2 I=\pi^{2} \times 2 \Rightarrow I=\pi^{2}\)
JEE Main-2023-06.04.2023
Integral Calculus
86758
\(\int_{0}^{\frac{\pi}{2}} \sin ^{5}\left(\frac{x}{2}\right) \cdot \sin x d x=\)
