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人気のある微分積分 >

integral of ((1+sin^2(x))/(1+cos(2x)))

解

\int (\frac{1 + sin ^{2} ( x )}{1 + cos ( 2 x )}) d x

解

\frac{1}{2} (2 tan (x) - x) + C

解答ステップ

\int \frac{1 + sin ^{2} ( x )}{1 + cos ( 2 x )} d x

三角関数の公式を使用して書き換える

= \int \frac{1 + sin ^{2} ( x )}{2 cos ^{2} ( x )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1 + sin ^{2} ( x )}{cos ^{2} ( x )} d x

拡張

\frac{1 + sin ^{2} ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

= \frac{1}{2} \cdot \int \frac{1}{cos ^{2} ( x )} + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{1}{cos ^{2} ( x )} d x + \int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x)

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

\int \frac{sin ^{2} ( x )}{cos ^{2} ( x )} d x = tan (x) - x

= \frac{1}{2} (tan (x) + tan (x) - x)

簡素化

\frac{1}{2} (tan (x) + tan (x) - x) : \frac{1}{2} (2 tan (x) - x)

= \frac{1}{2} (2 tan (x) - x)

定数を解答に追加する

= \frac{1}{2} (2 tan (x) - x) + C

グラフ

人気の例

integral of x^3-2x^2-x+2

\int x^{3} - 2 x^{2} - x + 2 d x

integral of-0.4x+20

\int - 0.4 x + 20 d x

integral of (3-x)/(1+x^2)

\int \frac{3 - x}{1 + x ^{2}} d x

integral of xsqrt(5x^2+3)

\int x 5 x^{2} + 3 d x

integral of (ln(y))/(y^2)

\int \frac{ln ( y )}{y ^{2}} d y