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Integral test Quiz 2

mathematics

 If $A = \int_1^{16} {\dfrac{{\sqrt x }}{{\sqrt {17 - x}  + \sqrt x }}dx} $, then write the value of $\dfrac{A}{2}$.

 Use the property $\int_a^b {f(x)} dx = \int_a^b {f(a + b - x)} dx$ to find the given integral.  

Consider the given integral: $ \\ A = \int_1^{16} {\dfrac{{\sqrt x }}{{\sqrt {17 - x}  + \sqrt x }}dx}  ---- \left( i \right) \\ $ Applying the property $\int_a^b {f\left( x \right)dx = } \int_a^b {f\left( {a + b - x} \right)dx} $ we get: $ \\ A = \int_1^{16} {\dfrac{{\sqrt {17 - x} }}{{\sqrt x  + \sqrt {17 - x} }}} dx ---- \left( {ii} \right) \\ $ Adding equation $\left( {\text{i}} \right)$and$\left( {{\text{ii}}} \right)$ we get: $ \\ 2A = \int_1^{16} {\dfrac{{\sqrt x  + \sqrt {17 - x} }}{{\sqrt x  + \sqrt {17 - x} }}} dx = \int_1^{16} 1 dx \\   \Rightarrow 2A = \left[ x \right]_1^{16} = 15 \\   \Rightarrow A = \dfrac{{15}}{2} \\   \Rightarrow \dfrac{A}{2} = \dfrac{{15}}{4} \\  $

 Find the value of $I = \int_0^{\frac{\pi }{2}} {\dfrac{{{{\sin }^{\frac{{15}}{{17}}}}x}}{{{{\sin }^{\frac{{15}}{{17}}}}x + {{\cos }^{\frac{{15}}{{17}}}}x}}dx} $

 Use property of definite integral $\int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} } $ to getting integral equation and add this equation to given equation and solve, then integrate.

Consider the given integral: $ \\ I = \int_0^{\frac{\pi }{2}} {\dfrac{{{{\sin }^{\frac{{15}}{{17}}}}x}}{{{{\sin }^{\frac{{15}}{{17}}}}x + {{\cos }^{\frac{{15}}{{17}}}}x}}dx} {\text{ ---- }}\left( {\text{i}} \right) \\ $ Use the property $\int_0^a {f\left( x \right)dx}  = \int_0^a {f\left( {a - x} \right)dx} $i.e. $ \\  I = \int_0^{\frac{\pi }{2}} {\dfrac{{{{\sin }^{\frac{{15}}{{17}}}}\left( {\dfrac{\pi }{2} - x} \right)}}{{{{\sin }^{\frac{{15}}{{17}}}}\left( {\dfrac{\pi }{2} - x} \right) + {{\cos }^{\frac{{15}}{{17}}}}\left( {\dfrac{\pi }{2} - x} \right)}}dx}  \\   \Rightarrow I = \int_0^{\frac{\pi }{2}} {\dfrac{{{{\cos }^{\frac{{15}}{{17}}}}x}}{{{{\cos }^{\frac{{15}}{{17}}}}x + si{n^{\frac{{15}}{{17}}}}x}}} dx{\text{ ---- }}\left( {{\text{ii}}} \right) \\  $ Add equations $\left( {\text{i}} \right)$and$\left( {{\text{ii}}} \right)$ we get: $ \\ 2I = \int_0^{\frac{\pi }{2}} {\dfrac{{si{n^{\frac{{15}}{{17}}}}x + {{\cos }^{\frac{{15}}{{17}}}}x}}{{si{n^{\frac{{15}}{{17}}}}x + {{\cos }^{\frac{{15}}{{17}}}}x}}} dx = \int_0^{\frac{\pi }{2}} 1 \,dx \\ $ Further solving the limits we get: $ \\ 2I = \left[ x \right]_0^{\frac{\pi }{2}} = \dfrac{\pi }{2} \\   \Rightarrow I = \dfrac{\pi }{4} \\  $

 Write the value of $\int_0^{\frac{\pi }{2}} {\sqrt {1 - \cos 6x} } dx$.

 First solve $1 - \cos 6x$ by using half angle formula then further solve integral.

Let, $  I = \int_0^{\frac{\pi }{2}} {\sqrt {1 - \cos 6x} } dx \\   \Rightarrow I = \int_0^{\frac{\pi }{2}} {\sqrt {2si{n^2}3x} } dx \\   \Rightarrow I = \sqrt 2 \int_0^{\frac{\pi }{2}} {\sin 3x} dx \\  $ Further solving we get: $ \\ \sqrt 2 \left[ { - \dfrac{{\cos 3x}}{3}} \right]_0^{\frac{\pi }{2}} \\   \Rightarrow \sqrt 2 \left[ {\dfrac{{ - \cos \dfrac{{3\pi }}{2}}}{3} + \dfrac{{\cos 0}}{3}} \right] \\   \Rightarrow \sqrt 2 \left( {0 + \dfrac{1}{3}} \right) = \dfrac{{\sqrt 2 }}{3} \\   \Rightarrow I = \dfrac{{\sqrt 2 }}{3} \\  $

 If $y = \int_0^{\frac{\pi }{2}} {xsinx} dx + \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {ta{n^5}x} dx$, then write value of $y$.

 Solve the two integrals separately and then add the two results. In the first integral, use integration by part and in the second integral show that the function is odd. 

Consider the given integral: $ \\ y = \int_0^{\frac{\pi }{2}} {xsinx} dx + \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {ta{n^5}x} dx \\ $ Consider, ${\text{ }}{I_1} = \int_0^{\frac{\pi }{2}} x \sin xdx$ $ \\ $ Using integration by parts we get: $ \\ {I_1} = \left[ {x\int {\sin } x\,dx - \int {\left( {\dfrac{d}{{dx}}(x)\int {\sin \,x} \,dx} \right)} \,dx} \right]_0^{\frac{\pi }{2}} \\   \Rightarrow {I_1} = \left[ { - x\cos x - \int {\left( {1 \times \left( { - \cos x} \right)} \right)dx} } \right]_0^{\frac{\pi }{2}} \\   \Rightarrow {I_1} = \left[ { - x\cos x + \int {\left( {\cos x} \right)dx} } \right]_0^{\frac{\pi }{2}} \\   \Rightarrow {I_1} = \left[ { - x\cos x + \sin x} \right]_0^{\frac{\pi }{2}} \\   \Rightarrow {I_1} = \left( { - \dfrac{\pi }{2}\cos \dfrac{\pi }{2} + \sin \dfrac{\pi }{2}} \right) - \left( {0 - \sin 0} \right) \\   \Rightarrow {I_1} = 1 \\  $ Consider, ${I_2} = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {{{\tan }^5}x} dx$ $ \\ $ Let $f(x) = {\tan ^5}x$ Now replacing $x$ by $ - x$ we get: $ \\ f\left( { - x} \right) = {\tan ^5}\left( { - x} \right) \\   \Rightarrow f =  - {\tan ^5}x \\   \Rightarrow f =  - f(x) \\  $ So $f(x)\,$ is an odd function. $ \\ {I_2} = \int_{ - \frac{\pi }{4}}^{\frac{\pi }{4}} {{{\tan }^5}x} \,dx = 0 \\ $ Further solving we get: $ \\ y = {I_1} + {I_2} \\   \Rightarrow y = 1 + 0 \\   \Rightarrow y = 1 \\  $

 Find the value of the integral $\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\dfrac{{\pi dx}}{{{e^{\sin x}} + 1}}} $

 Use property of definite integral $\int_a^b {f\left( x \right)} dx = \int_a^b {f\left( {a + b - x} \right)} dx$ to get integral equation and add this equation the given equation and solve the integral.

Consider the given integral: $ \\ I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi dx}}{{{e^{\sin x}} + 1}}} {\text{ ---- }}\left( {\text{i}} \right) \\ $ Using property $\int_a^b {f\left( x \right)} dx = \int_a^b {f\left( {a + b - x} \right)} dx$ we get: $ \\   I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi \,dx}}{{{e^{\sin \left( {0 - x} \right)}} + 1}}}  = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi \,dx}}{{{e^{ - \operatorname{sinx} }} + 1}}}  \\   \Rightarrow I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi {e^{\sin x}}\,}}{{{e^{\operatorname{sinx} }} + 1}}} dx{\text{ ---- }}\left( {{\text{ii}}} \right) \\  $ Adding equation$\left( {\text{i}} \right)$ and$\left( {{\text{ii}}} \right)$ we get: $ \\   2I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi \,}}{{{e^{\operatorname{sinx} }} + 1}}} dx + \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi {e^{\operatorname{sinx} }}\,}}{{{e^{\operatorname{sinx} }} + 1}}dx}  \\   \Rightarrow 2I = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\dfrac{{\pi \left( {{e^{\sin x}} + 1} \right)\,}}{{{e^{\operatorname{sinx} }} + 1}}} dx = \pi \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {1\,} dx \\   \Rightarrow 2I = \pi \left[ x \right]_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} = {\pi ^2}I = \dfrac{{{\pi ^2}}}{2} \\  $