0-2-sinx-sinx-cosx-dx-

Question Number 7300 by Tawakalitu. last updated on 22/Aug/16

\int_{0}^{\frac{Π}{2}} \frac{s i n x}{s i n x + c o s x} d x

Answered by Yozzia last updated on 22/Aug/16

I = \int_{0}^{π / 2} \frac{s i n x}{s i n x + c o s x} d x

Q = \int_{0}^{π / 2} \frac{c o s x}{s i n x + c o s x} d x

I + Q = \int_{0}^{π / 2} \frac{s i n x + c o s x}{s i n x + c o s x} d x = \int_{0}^{π / 2} d x = \frac{π}{2}

Q - I = \int_{0}^{π / 2} \frac{c o s x - s i n x}{s i n x + c o s x} d x = \int_{1}^{1} \frac{1}{u} d u = 0

\Rightarrow I = Q ∴ 2 I = \frac{π}{2} \Rightarrow I = \frac{π}{4} .

Commented by Yozzia last updated on 22/Aug/16

I (m) = \int_{0}^{π / 2} \frac{{s i n}^{m} x}{{c o s}^{m} x + {s i n}^{m} x} d x m \in R .

∵ \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

\Rightarrow I (m) = \int_{0}^{π / 2} \frac{{s i n}^{m} (\frac{π}{2} - x)}{{c o s}^{m} (\frac{π}{2} - x) + {s i n}^{m} (\frac{π}{2} - x)} d x

S i n c e s i n (0.5 π - x) = c o s x a n d c o s (0.5 π - x) = s i n x

\Rightarrow I (m) = \int_{0}^{π / 2} \frac{{c o s}^{m} x}{{s i n}^{m} x + {c o s}^{m} x} d x .

∴ 2 I (m) = \int_{0}^{π / 2} \frac{{s i n}^{m} x + {c o s}^{m} x}{{s i n}^{m} x + {c o s}^{m} x} d x = \frac{π}{2}

\Rightarrow I (m) = \frac{π}{4} o r \int_{0}^{π / 2} \frac{{s i n}^{m} x}{{c o s}^{m} x + {s i n}^{m} x} d x = \frac{π}{4} .

Commented by Tawakalitu. last updated on 22/Aug/16

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