What is (i1)x^sinx+(sin(x))^cosx ?

The solution to (i1)x^sinx+(sin(x))^cosx is i^2x^{s+1}n+sin^c(x)osx

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify (i1)x^sinx+(sin(x))^cosx

simplify

(i 1) x^{s} in x + (sin (x))^{c} os x

i^{2} x^{s + 1} n + sin^{c} (x) os x

Solution steps

(i \cdot 1) x^{s} in x + sin^{c} (x) os x

Apply rule:

a \cdot 1 = a

(i \cdot 1) x^{s} in x = i x^{s} in x

= i x^{s} in x + sin^{c} (x) os x

Apply exponent rule:

aa = a^{2}

ii = i^{2}

= i^{2} x^{s} n x + sin^{c} (x) os x

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

x^{s} x = x^{s + 1}

= i^{2} x^{s + 1} n + sin^{c} (x) os x

(\frac{( 2 x )}{5 + x ^{3}})^{2}

- 10 x (x - 6)

(x - 5)^{5} (- 3)

- 4 (5 y + 7)

x^{8} - 49 y^{6}

What is (i1)x^sinx+(sin(x))^cosx ?
The solution to (i1)x^sinx+(sin(x))^cosx is i^2x^{s+1}n+sin^c(x)osx